THEORY   OF   STRUCTURES 


AND 


STRENGTH  OF  MATERIALS. 


WITH 


DIAGRAMS,    ILLUSTRATIONS,    AND   EXAMPLES. 


BY 

HENRY  T.  BOVEY,  M.A.,  D.C.L.,  F.R.S.C., 

PROFESSOR    OF   CIVIL   ENGINEERING   AND   APPLIED    MECHANICS,    M*GILL    UNIVERSITY,    MONTREAL. 
MEMBER   OF   THE   INSTITUTION    OF    CIVIL   ENGINEERS;     MEMBER   OF   THE   INSTITU- 
TION  OF   MECHANICAL    ENGINEERS;     LATE    FELLOW   OP 

QUEENS'  COLLEGE,  CAMBRIDGE  (ENG.). 


FIRST   THOUSAND. 


NEW  YORK: 

JOHN     WILEY    &     SONS, 

53  EAST  TENTH  STREET. 

1893. 


COPYRIGH^,  1893, 

BV 
HENRY  T.  BOVEY. 


ROBERT  DRTJMMOND, 


886  Pearl  Street, 
New  York. 


444  &  446  Pearl  Street, 
New  York. 


DEDICATED 


WHOSE   BENEFACTIONS   TO   M*GILL  UNIVERSITY 

HAVE  DONE   SO   MUCH  TO  ADVANCE   THE   CAUSE  OF 

SCIENTIFIC  EDUCATION. 


PREFACE. 


THE  present  work  treats  of  that  portion  of  Applied 
Mechanics  which  has  to  do  with  the  Design  of  Structures. 

Free  reference  has  been  made  to  the  works  of  other 
authors,  yet  a  considerable  amount  of  new  matter  has  been 
introduced,  as,  for  example,  the  Articles  on  "  Surface  Loading" 
by  Carus-Wilson,  "  The  Flexure  of  Columns  "  by  Findlay,  and 
"  The  Efficiency  of  Riveted  Joints  "  by  Nicolson ;  also  my 
own  Articles  on  "  Maximum  Shearing  Forces  and  Bending 
Moments,"  "  The  Flexure  of  Long  Columns,"  "  The  Theorem 
of  Three  Moments,"  etc. 

I  am  much  indebted  to  Messrs.  C.  F.  Findlay  and  W.  B. 
Dawson  for  valuable  information  respecting  the  treatment  of 
Cantilever  Bridges,  Arched  Ribs,  and  the  Live  Loads  on  Bridges. 

To  Messrs.  J.  M.  Wilson,  P.  A.  Peterson,  C.  Macdonald, 
and  others,  many  thanks  are  due  for  data  respecting  the  Dead 
Weights  of  Bridges. 

I  am  under  deep  obligation  to  my  friend  Prof.  Chandler, 
who  has  kindly  revised  the  proof-sheets,  and  who  has  made 
many  important  suggestions. 

I  have  endeavored  so  to  arrange  the  matter  that  the 
student  may  omit  the  advanced  portions  and  obtain  a  com- 
plete elementary  course  in  natural  sequence. 

At   the   end    of   each    chapter,    a    number  of    Examples, 


VI  PREFACE. 

selected  for  the  most  part  from  my  own  experience,  are 
arranged  with  a  view  to  illustrating  the  subject-matter — an 
important  feature,  as  it  is  admitted  that  the  student  who  care- 
fully works  out  examples  obtains  a  mastery  of  the  subject 
which  is  otherwise  impossible. 

The  various  Tables  in  the  volume  have  been  prepared  from 
the  most  recent  and  reliable  results. 

A  few  years  ago  I  published  a  work  on  "  Applied  Me- 
chanics," consisting  mainly  of  a  collection  of  notes  intended 
for  the  use  of  my  own  students.  The  present  volume  may  be 
considered  as  a  second  edition  of  that  work,  but  the  subject- 
matter  has  been  so  much  added  to  and  rearranged  as  to  make 
it  almost  a  new  book.  I  venture  to  hope  that  this  volume 
may  prove  acceptable  not  only  to  students,  but  to  the  profes- 
sion at  large. 

HENRY  T.  BOVEY. 

McGiLL  COLLEGE,  MONTREAL, 
November,  1892. 


CONTENTS. 


CHAPTER  I. 

FRAMED  STRUCTURES. 

PAGE 

Definitions I 

Frames  of  Two  or  More  Members 2 

Funicular  Polygon 3 

Polygon  of  Forces 4 

Line  of  Loads 5 

Mansard  Roof 6 

Non-closing   Polygons ...  7 

Funicular  Curve 10 

Centre  of  Gravity 1 1 

Moment  of  Inertia 12 

Cranes,  Jib 13 

"        Derrick 16 

"        Composite 31 

Shear  Legs ^ \~f 

Bridge  Trusses 17 

Roof  Trusses i  > 

King-post  Truss 21 

Incomplete  Frames 27 

Queen-post  Truss 31 

Composite  Frames 32 

Roof-weights 37 

Wind-pressure 38 

Distribution  of  Loads 39 

Examples  of  Roof  Trusses 41 

Examples  of  Bridge   Trusses  (Fink,  Bollman,  Howe,    Bowstring,    Single- 
intersection,  etc) 52 

Method  of  Sections 62 

Piers 65 

vii 


VI  il  CONTENTS. 


Tables  of  Roof  -weights  and  Wind-pressures  ............................     67 

Examples  .............................................................     7° 

CHAPTER  II. 
SHEARING  FORCES  AND  BENDING  MOMENTS. 

Equilibrium  of  Beams  ................................................  93 

Shearing  Force  ..................................  ......................  95 

Bending  Moment  ...................    ................................  96 

Examples  of  Shearing  Force  and  Bending  Moment  ......................  97 

Relation  between  Shearing  Force  and  Bending  Moment  ..................  108 

Effect  of  Live  (or  Rolling)  Load  ........................................  in 

Graphical  Representation  of  Moment  of  Forces  with  Respect  to  a  Point..  .  116 

Relation  between  Bending  Moments  and  Funicular  Polygon  ..............  118 

Maximum  Shear  and  Maximum  Bending  Moment  at  any  Point  of  an  Arbi- 

trarily Loaded  Girder  .............................................  121 

Hinged  Girders  .......................................................  126 

Examples  ...........  '  ................................................  131 

CHAPTER  III. 
GENERAL  PRINCIPLES,  ETC. 

Definitions.  .  .  «  .......................................  ................  140 

Stress,  Simple  ........................................................  140 

"       Compound  ....................................................  140 

Hooke's  Law  ........................................................  141 

Coefficient  of  Elasticity  ...................................  ............  141 

Poisson's  Ratio  ........................  .  ..............................  142 

Effect  of  Temperature  .................................................  142 

Specific  Weight  .......................................................  143 

Limit  of  Elasticity  ...................................................  143 

Breaking  Stress  .......  ...............................................  147 

Dead  and  Live  Loads  .................................................  143 

Repeated  Stress  Effect  ................................................  145 

Wohler's  Experiments  ................................................  145 

Testing  of  Metals  ....................................................  147 

Launhardt's  Formula  ..................................................  159 

Wey  ranch's  Formula  ...............  ...................................  153 

Unwin's  Formula  .........................................  .  .........  159 

Flow  of  Solids  .......................................................  162 

Work,  Internal  and  External  ...........................................  168 

Energy,  Kinetic  and  Potential  .........................................  167 

Oblique  Resistance  ..................................................  169 

Values  of  k  ..........................................................  174 

Momentum.        Impulse  ..................  .  ............................  176 

Angular  Momentum  ..................................................  177 


CONTENTS. .  IX 

PAGE 

Useful  Work.     Waste  Work 178 

Centrifugal  Force 181 

I  m  pact 1 84 

Extension  of  a  Prismatic  Bar 189 

Oscillatory  Motion  of  a  Weight  at  the  End  of  a  Vertical  Elastic  Rod 190 

Inertia  198 

Balancing 198 

Curves  of  Piston  Velocity 205 

Linear  Diagrams  of  Velocity 206 

Curves  of  Crank-effort. . . 207 

Curves  of  Energy 207 

Fluctuation  of  Energy 207 

Tables  of  Strengths,  Elasticities,  and  Weights  of  Materials 210 

Tables  of  the  Breaking  Weights  and  Coefficients  of  Bending  Strengths  of 

Beams 213 

Table  of  the  Weights  and  Crushing  Weights  of  Rocks,  etc 214 

Table  of  Expansions  of  Solids 215 

Examples 216 


CHAPTER  IV. 

STRESSES,  STRAINS,  EARTHWORK,  AND  RETAINING  WALLS. 

Internal  Stresses 235 

Simple  Strain 235 

Compound  Strain 236 

Principal  Stresses 240 

Curves  of  Maximum  Shear  and  Normal  Intensity 240 

Combined  Bending  and  Twisting  Stresses 244 

Combined  Longitudinal  and  Twisting  Stresses 247 

Conjugate   Stresses 247 

Relation  between  Principal  and  Conjugate  Stresses 247 

Ratio  of  Conjugate  Stresses 250 

Relation  between  Stress  and  Strain 251 

Rankine's  Earthwork  Theory 255 

Pressure  against  a  Vertical  Plane 257 

Earth  Foundations 258 

Retaining  Walls 260 

Retaining  Walls.     Conditions  of  Equilibrium 260 

Rankine's  Earthwork  Theory  applied  to  Retaining  Wall? 264 

Line  of  Rupture   265 

Practical  Rules  respecting  Retaining  Walls 267 

Reservoir  Walls 271 

General  Case  of  Reservoir  Walls 275 

General  Equations  of  Stress 276 

Ellipsoid  of  Stress 281 

Stress-strain  Equations 281 


X  CONTENTS. 

PAGE 

Isotropic  Bodies 283 

Relation  between  A ,  A,  and  G 285 

Traction 287 

Torsion 288 

Work  done  in  the  Small  Strain  of  a  Body  (Clapeyron) 292 

Examples 294 


CHAPTER  V. 
FRICTION. 

Friction 300 

Laws  of  Friction 300 

Inclined  Plane 301 

Wedge , 302 

Screws . 306 

Endless  Screw 309 

Rolling  Friction 310 

Journal  Friction 312 

Pivot 316 

Cylindrical  Pivot 316 

Wear 318 

Conical  Pivot 319 

Schiele's  Pivot  {anti-friction) 320 

Belts  and  Ropes 321 

Brakes 323 

Effective  Tension  of  a  Belt 324 

Effect  of  High  Speed 325 

Slip  of  Belts 326 

Prony's  Dynamometer 327 

Stiffness  of  Belts  and  Ropes 327 

Wheel  and  Axle 329 

Toothed  Gearing 331 

Bevel- wheels 335 

Efficiency  of  Mechanisms 335 

Table  of  Coefficients  of  Journal  Friction 336 

Examples 337 

CHAPTER  VI. 

t 

TRANSVERSE  STRENGTH  OF  BEAMS. 

Elastic  Moment 340 

Moment  of  Resistance 340 

Neutral  Axis 340 

Transverse  Deformation 344 

Coefficient  of  Bending  Strength 344 


CONTENTS.  XI 

PAGE 

Equalization  of  Stress 349 

Surface  Loading 350 

Effect  of  Bending  Moment  in  a  Plane  which  is  not  a  Principal  Plane 354 

Springs 355 

Beams  of  Uniform  Strength 358 

Flanged  Girders 365 

Classification  of  Flanged  Girders 365 

Equilibrium  of  Flanged  Girders , 366 

Moments  of  Inertia  of  I  and  other  Sections 371 

Design  of  a  Girder  of   I-section 381 

Deflection  of  Girders 384 

Camber 387 

Stiffness 389 

Distribution  of  Shearing  Stress 391 

Beam  acted  upon  by  Forces  Oblique  to  its  Direction 396 

Similar  Girders 401 

Allowance  to  be  made  for  Weight  of  Beam 405 

Examples 407 


CHAPTER  VII. 
TRANSVERSE  STRENGTH  OF  BEAMS — {Continued.} 

General  Equations 428 

Interpretation  of  the  General  Equations 432 

Examples— Cantilever 435 

Girder  upon  Two  Supports 439 

"      fixed  at   One  End   and  resting   upon    Support    at  the 

Other 442 

"                 "     fixed  at  Both  Ends 445 

"                  "     upon  Two  Supports  not  in  the  same  Horizontal  Plane.  446 

"            Neutral  Axis  of  Arbitrarily  Loaded  Girders 448 

"            Cantilever  with  Varying  Section 455 

"            Girder  Encastre  at  the  Ends 458 

Springs 456 

Work  done  in  bending  a  Beam 460 

Transverse  Vibrations  of  a  Beam  supported  at  Both  Ends ,  461 

Imperfect  Fixture 461 

Continuous  Girders 463 

Theorem  of  Three  Moments 463 

Swing-bridges 470 

Maximum   Bending  Moment  at  the  Points  of   Support  of  a   Continuous 

Girder  of  n  Spans 475 

General  Theorem  of  Three   Moments 484 

Comparative  Merits  of  Continuous  Girders 486 

Examples 490 


Xll  CONTENTS. 

CHAPTER  VIII. 
PILLARS. 

PAGE 

Classification  of  Pillars 513 

Form  "        "      514 

Failure  "         "      515 

Uniform  Stress 516 

Uniformly  Varying  Stress  517 

Hodgkinson's  Formulae 520 

Gordon's  Formula 522 

Values  of  the  Coefficients  (a  and  f)  in  Gordon's  Formula 523 

Graphical  Representation  of  Strength  of  Pillars 524 

Rankine's  Modification  of  Gordon's  Formula 526 

Formula  for  Safe  Working  Stress 526 

Value  of  "  Radius  of  Gyration  "  for  Different  Sections 526 

American  Iron  Columns 532 

Long  Thin  Pillars 534 

Long  Columns  of  Uniform  Section  (Euler's  Theory) 538 

Resistance  of  Columns  to  Buckling  (Weyrauch's  Theory) 550 

Baker's  Formulae ....    549 

Flexure  of  Columns , 554,  557 

Examples 563 

CHAPTER  IX. 

TORSION. 

Definition 568 

Coulomb's  Laws 568 

Torsional  Strength  of  Shafting 569 

St.  Venant's  Results 572 

Torsional  Rupture 572 

Resilience  of   Shafting 574 

Effect  of  Combined  Bending  and  Twisting 574 

Distance  between  Bearings  for  Shafting 575 

Efficiency  of  Shafting 577 

Spiral  Springs 577 

Figures  illustrating   the  Distortion   produced  by  twisting   Round,   Square, 

and  Rectangular  Iron  Bars 5850,  585^ 

Examples 580 

CHAPTER  X. 
CYLINDRICAL  AND  SPHERICAL  BOILERS. 

Cylinders 586 

Efficiency  of  Riveted  Joints  in  Boilers 587 


CONTENTS.  X1U 

PAGE 

Thick  Hollow  Cylinder 588 

Spherical  Shells 591 

Practical  Formulae 592 

Examples 594 

CHAPTER  XL 
BRIDGES. 

Classification 597 

Curved  and  Horizontal  Flanges 597 

Depth  of  Girders  (or  Trusses) 597 

Position  of  Platform 598 

Comparative  Advantages  of  Two,  Three,  and  Four  Main  Girders 600 

Dead  Load 600 

Live  Load 600 

Trellis  (or  Lattice)  Girder 600 

Warren  Girder 603 

Howe  Truss 61 1 

Single-intersection  Truss 616 

Double-intersection  Truss 616 

Whipple  Truss 618 

Linvillc  Truss 618 

Post  Truss 618 

Quadrangular  Truss 618 

Bowstring  Truss 618 

Bowstring  Truss  with  Isosceles  Bracing 624 

Bowstring  Suspension-bridge   (Lenticular  Truss) 626 

Cantilever  Trusses 627 

Curve  of  Cantilever  Boom 634 

Deflection  of  Cantilevers 638 

Rollers 639 

Wind-pressure 651 

Regulations  respecting  Wind-pressure 653 

Lateral  Bracing 654 

Chords 655 

Stringers 656 

Maximum  Allowable  Working  Stresses 657 

Camber 659 

Rivet  Connections  between  Flange  and  Web 660 

Eye-bars,  Pins,  and   Rivets 661 

Steel  Eyebars .   665 

Rivets 666 

Dimensions   of  Rivets 667 

Strength  of  Punched  and  Drilled  Plates 668 

Riveted  Joints 668 

Theory  of  Riveted  Joints 671 


XIV  CONTENTS. 


Covers 675 

Efficiency  of  Riveted  Joints 676 

Tables  of  Weights  of  Actual  Bridges 682-687 

Table  of  Loads  for  Highway  Bridges 688 

Examples 689-702 

CHAPTER  XII. 

SUSPENSION-BRIDGES. 

Cables 703 

Anchorage 704 

Suspenders . .  706 

Curve  of  Cable  (catenary) 706 

Link  "  Cable" 709 

Length  of  Cable 712 

Weight  of  Cable 713 

Deflection  due  to  Change  of  Length 714 

Pressure  upon  Piers 718 

Stiffening  Truss 719 

Stiffening  Truss  hinged  at  the  Centre 725 

Suspension-bridge  Loads 730 

Modifications  of  the  Suspension-bridge  proper 731 

Examples 734~739 

CHAPTER  XIII. 
ARCHED  RIBS. 

Definitions 740 

Equilibrated  Polygon  (Line  of  Resistance) 741 

Polygon  of  Pressures 743 

Linear  Arch 743 

Conditions  of  Equilibrium 745 

Joint  of  Rupture 747 

Minimum  Thickness  of  Abutment 749 

Empirical  Formulae 750 

Linear  Arch  in  Form  of  a  Parabola 750 

"         "     "        "        "     Transformed  Catenary 750 

"          "     "        "        "     Circular  Arc 753 

"          "     "        "        "     Elliptic  Arc 753 

Hydrostatic  Arch 757 

Geostatic  Arch , 759 

General  Arch  Theory 760 

Arched   Ribs   762 

Bending  Moment  and  Thrust  at  any  Point  of  an  Arched  Rib 763 

Rib  with  Hinged  Ends 764 


CONTENTS.  XV 

PAGE 

Semicircular  Rib  with  Hinged  Ends ' 765 

Graphical  Determination  of  the  Thrust  at  any  Point  of  an  Arched  Rib 767 

Rib  in  Form  of  a  Circular  Arc 769 

Rib  with  Fixed  Ends * 771 

"      "         "         "    in  Form  of  a  Circular  Arc 773 

"      "         "         "     "       «'         "     Semicircle 775 

"      "         "         "     "       "         "     Parabola 775 

Effect  of  a  Change  of  Temperature 777,  786 

Deflection  of  an  Arched  Rib 780,  802 

Elementary  Deformation  of  an  Arched  Rib 781 

Rib  of  Uniform  Stiffness 788 

Parabolic  Rib  of  Uniform  Depth  and  Stiffness 789,  795,  800 

Arched  Rib  of  Uniform  Stiffness  with  Fixed  Ends 804 

Stresses  in  Spandril  Posts  and  Diagonals 804 

Maxwell's  Method  of  Determining  the  Stresses  in  a  Framed  Arch 806 

Examples 809-812 


THEORY   OF   STRUCTURES. 


CHAPTER   I. 
FRAMES   LOADED   AT   THE  JOINTS. 

I.  Definitions. — Frames  are  rigid  structures  composed  oi 
straight  struts  and  ties,  jointed  together  by  means  of  bolts, 
straps,  mortises  and  tenons,  etc.  Struts  are  members  in  com- 
pression, ties  members  in  tension,  and  the  term  brace  is  applied; 
to  either. 

The  external  forces  upon  a  frame  are  the  loads  and  the 
reactions  at  the  points  of  support,  from  which  may  be  found 
the  resultant  forces  at  the  joints.  The  greatest  care  should  be 
exercised  in  the  design  of  the  joints.  The  resultant  forces 
should  severally  coincide  in  direction  with  the  axes  of  the 
members  upon  which  they  act,  and  should  intersect  the  joints 
in  their  centres  of  gravity.  Owing  to  a  want  of  homogeneity 
in  the  material,  errors  of  workmanship,  etc.,  this  coincidence  is 
not  always  practicable,  but  it  should  be  remembered  that  the 
smallest  deviation  introduces  a  bending  action.  Such  an 
action  will  also  be  caused  by  joint  friction  when  the  frame  is 
insufficiently  braced.  The  points  in  which  the  lines  of  action 
of  the  resultants  intersect  the  joints  are  also  called  the  centres 
of  resistance,  and  the  figure  formed  by  joining  the  centres  of 
resistance  in  order  is  usually  a  polygon,  which  is  designated  the 
line  of  resistance  of  the  frame. 

The  position  of  the  centres  should  on  no  account  be  allowed 
to  vary.  It  is  assumed,  and  is  practically  true,  that  the  joints 
of  a  frame  are  flexible,  and  that  the  frame  under  a  given  load 


2  THEORY  OF  STRUCTURES. 

i 

•does  not  sensibly  change  in  form.  Thus  an  individual  mem. 
her  is  merely  stretched  or  compressed  in  the  direction  of  its 
length,  i.e.,  along  its  line  of  resistance,  while  the  frame  as  a 
whole  may  be  subjected  to  a  bending  action. 

The  term  truss  is  often  applied   to  a  frame  supporting  a 
weight, 

2.  Frame   of  Two  Members.  —  OA,    OB   are   two   bars 
jointed  at  O  and  supported  at  the  ends  A,  B.     The  frame  in 


FIG. 


FIG.  2. 


FIG.  3. 


Fig.  I  consists  of  two  ties,  in  Fig.  3  of  two  struts,  and  in  Fig. 
2  of  a  strut  and  a  tie. 

Let  P  be  the  resultant  force  at  the  joint,  and  let  it  act  in 
the  direction  OC.  Take  OC  equal  to  P  in  magnitude,  and 
draw  CD  parallel  to  OB.  OD  is  the  stress  along  OA,  and  CD 
is  that  along  OB. 

Let  the  angle  AOB  —  a,  and  the  angle  COD  =  /3. 

Let  S, ,  52  be  the  stresses  along  OA,  O£,  respectively. 


S,  _  OD  _  sin(<x  —  ft) 
~P  ~~OC~        sin  a~ 


,     S,  _  CD  _  sin  ft 
P  ~  OC  ~  sin  a 


3.  Frame  of  Three  or  More  Members- — Let  A.A^A^ . . . 

be  a  polygonal  frame  jointed  at  Al9AttA91  .  .  .  Let  Pl , 
/*,,  P3,  .  .  .  be  the  resultant  forces  at  the  joints  Alt  A9,  A3, 
.  .  .  ,  respectively.  Let  Slt  Sa,  53,  .  .  .  be  the  forces  along 
AtA9t  A^AZ,  .  .  .  ,  respectively. 

Consider  the  joint  A^. 

The  lines  of  action  of  three  forces,  Plt  Sl ,  and  S9,  intersect 
in  this  joint,  and  the  forces,  being  in  equilibrium,  may  be 
represented  in  direction  and  magnitude  by  the  sides  of  the 


FRAMES  LOADED  AT   THE  JOINTS.  3 

triangle  Os,s6,  in  which  v6  is  parallel  to  Plt  OsJ  to  Slt  and  Os9 
to  5.. 

Similarly,/^,  St ,  52  maybe  represented  by  the  sides  of 
the  triangle  Os^.,  which  has  one  side,  Osiy  common  to  the 
triangle  Os^^,  and  so  on. 

Thus  every  joint  furnishes  a  triangle  having  a  side  common 
to  each  of  the  two  adjacent  triangles,  and  all  the  triangles  to- 
gether form  a  closed  polygon  s^^.  .  .  The  sides  of  this 
polygon  represent  in  magnitude  and  direction  the  resultant 


FIG.  4. 


forces  at  the  joints,  and  the  radii  from  the  pole  O  to  the  angles 
jjVs  ».«•'•  represent  in  magnitude,  direction  and  character, 
the  forces  along  the  several  sides  of  the  frame  A^AZA^,  .  . 
The  polygon  A^A^A3  ...  is  the  line  of  resistance  of  the 
frame,  and  is  called  the  funicular  polygon  of  the  forces  Plt  P39 
/*,...  with  respect  to  the  pole  O. 

The  two  polygons  are  said  to  be  reciprocal,  and,  in  general, 
two  figures  in  graphical  statics  are  said  to  be  reciprocal  when 
the  sides  in  the  one  figure  are  parallel  or  perpendicular  to  cor- 
responding sides  in  the  other. 

A  triangle  or  polygon  is  also  said  to  be  the  reciprocal  of  a 
point  when  its  sides  are  parallel  or  perpendicular  to  correspond- 
ing lines  radiating  from  the  point.  Thus  the  triangle  Os^  is 


4  THEORY  OF  STRUCTURES. 

the  reciprocal  of  the  point  Alt  and  the  polygon 
is  the  reciprocal  of  the  point  O. 

If  more  than  two  members  meet  at  a  joint,  or  if  the  joint  is 
subjected  to  more  than  one  load,  the  resulting  force  diagram 
will  be  a  quadrilateral,  pentagon,  hexagon,  .  .  .  according  as 
the  number  of  members  is  3,  4,  5,  ...  or  the  number  of  loads 
2*  3»  4,  •  •  * 

In  practice  it  is  usually  required  to  determine  the  stresses 
in  a  number  of  members  radiating  from  a  joint  in  a  framed 
structure.  If  the  reciprocal  of  the  joint  can  be  drawn,  its 
sides  will  represent  in  direction  and  magnitude  the  stresses  in 
the  corresponding  members. 

Corollary. — The  converse  of  the  preceding  is  evidently  true. 
For  if  a  system  of  forces  is  in  equilibrium,  the  polygon  of 
forces  -V2.y3  .  .  .  must  close,  and  therefore  the  polygon  which 
has  its  sides  respectively  parallel  to  the  radii  from  a  pole  O  to 
the  angles  s1 ,  s2,  ss ,  .  .  .  and  which  has  its  angles  upon  the 
lines  of  action  of  the  forces,  must  also  close. 

EXAMPLE  I.  Let  O  be  a  joint  in  a  framed  structure,  and 
let  Os1}  Os^,  Oss,  ...  be  the  axes  of  the  members  radiating 
from  it.  The  polygon  A^A^A^  ...  is  the  reciprocal  of  O,  the 
side  A^A^  representing  the  stress  along  Oslt  the  side  A^A3  that 
along  Osz,  etc. 

Ex.  2.  Let  the  resultant  forces  at  the  joints  be  paral- 
lel. The  polygon  of  forces  becomes  the  straight  line  v5, 


FIG.  6. 


FIG.  7. 


FRAMES  LOADED  AT    THE  JOINTS.  5 

which  is  often  termed  the  line  of  loads.  Thus,  the  forces  Ps, 
/>„..../>,  are  represented  by  the  sides  s,s^  sys3,  .  .  .  v*>  which 
are  in  one  straight  line  closed  by  s^t  and  s&st,  representing  the 
remaining  forces  Pl  and  Pt,  and  the  triangles  Oststt  Os^3,  .  .  . 
are  the  reciprocals  of  the  points  At,  A,.  .  .  .  Draw  OH  per- 
pendicular to  stst.  The  projection  of  each  of  the  lines  Oslt 
Oft,  Os3,  .  .  .  perpendicular  to  s^  is  the  same  and  equal  to 
OH,  which  therefore  represents  in  magnitude  and  direction 
the  stress  which  is  the  same  for  each  member  of  the  frame. 

Let    alf  #2,  a3,  .  .  .  be   the   inclinations  of  the  members 
A,A^t  A^AS,  .  .  .  respectively,  to  the  line  of  loads.  Then 

OH  —  Hsl  tan  al  =  Hs&  tan  ab  ; 
.-.  Off  (cot  a  ;  -f  cot  a6)  =  ffs1  +  ffs,  —  s.s, 


and  OH,  in  direction  and  magnitude,  is  equal  to  the  stress 
common  to  each  member.  Also,  the  stress  in  any  member, 
e.g.,  A^AS  —  Os,  —  Offcosec  <xt  . 

Corollary.  —  Let  the  resultant  forces  at  the  joints  Alt  A6 
be  inclined  to  the  common  direction  of  the  remaining  forces, 
and  act  in  the  directions  shown  by  the  dotted  lines.  Let  P/, 
Pt'  be  the  magnitudes  of  the  new  forces  ;  draw  s^'  parallel  to 
the  direction  of  Pt'  so  as  to  meet  Os6  in  se'  ;  join  s^s^.  Since 
there  is  equilibrium,  s6's6  must  be  parallel  to  the  line  Sj 
of  action  of  Pt'  .  Thus,  s^s^  is  the  force  polygon. 

Ex.  3.  The  forces,  or  loads,  />,  Pt,  .  .  .  P6  are 
generally  vertical,  while  Plt  P6  are  the  vertical  re- 
actions of  the  two  supports. 

Suppose,  e.g.,  that  AtAt  .  .  .  A6  is  a  rope  or  chain 
suspended  from  the  points  Alt  A6,  in  a  horizontal 
plane  and  loaded  at  A^AS  .  .  .  with  weights  P9,  Ptt  .  .  . 
The  chain  will  hang  in  a  form  dependent  upon  the 
magnitude  of  these  weights.  The  points  H  and  S6 
will  coincide,  and  Off  will  represent  the  horizontal 

riG.   8. 

tension   of   the  chain. 

Let  the  polygon  A^A^  .  .  .  At  be  inverted,  and  let  the  rope 
be  replaced  by  rigid"  bars,  A.A^  A^AZ.  .  .     The  diagram  of 


O  THEORY  OF  STRUCTURES. 

forces  will  remain  the  same,  and  the  frame  will  be  in 
equilibrium  under  fat  given  loads.  The  equilibrium,  however, 
is  unstable  as  the  chain,  and  consequently  the  inverted  frame 
will  change  form  if  the  weights  vary.  Braces  must  then  be 
introduced  to  prevent  distortion. 


-H 


FIG.  9. 


FIG.  10. 


Tal^e  the  case  of  a  frame  DCBA  .  .  .  symmetrical  with 
respect  to  a  vertical  through  A,  and  let  the  weights  at  A,  B,  C, 
.  .  .  be  Wlt  W^  Wz,  .  .  .  ,  respectively. 

Drawing  the  stress  diagram  in  the  usual  manner,  OH  rep- 
resents the  horizontal  thrust  of  the  frame. 

The  portions  s^^,  s^s3  ,  ...  of  the  line  of  loads  give  a 
definite  relation  between  the  weights  for  which  the  truss  will 
be  stable.  The  result  may  be  expressed  analytically,  as 
follows  : 

Let  arj,  ora,  of,,  .  .  .  be  the  inclinations  of  AB,  BC,  CD,  .  .  .  , 
respectively,  to  the  horizontal. 

Let  the  horizontal  thrust  OH  =  H.  Then 


It  ^  =  IV,=  IV3  =  .  .. 

cot  af1  =  3  cot  or2  =  5  cot  ofa  =  .  .  . 

If  there  are  two  bars  only,  viz.,  AB,  BC,  on  each  side  of  the 
vertical  centre  line,  the  frame  will  have  a  double  slope,  and  in 
this  form  is  employed  to  support  a  Mansard  roof. 


FRAMES  LOADED   A  T    THE  JOINTS.  ? 

4.  Non-closing  Polygons. — Let  a  number  of  forces  P19 
P9,  P3,  .  .  .  act  upon  a  structure,  and  let  these  forces,  taken  in 
order,  be  represented  in  direction  and  magnitude  by  the  sides 
of  the  unclosed  figure  MNPQ  .  .  .  This  figure  is  the  unclosed 


polygon  of  forces,  and  its  closing  line  TM  represents  in  direction 
and  magnitude  the  resultant  of  the  forces  Pl ,  P2 ,  P3 ,  .  .  . 

For  PM  is  the  resultant   of  P^  and  P^  and   may  replace 
them  ;  QM  may  replace  PM  and  P3,  i.e.,  Plt.P9,  and  />,;  and 


so  on. 


Take  any  point  O  and  join  OM,  ON,  OP,  .  .  . 

Draw  a  line  AB  parallel  to  OM  and  intersecting  the  line  of 
action  of  Pl  in  any  point  B.  Through  B  draw  BC  parallel  to 
ON  and  cutting  the  line  of  action  of  P^  in  C.  Similarly,  draw 
CD  parallel  to  OP,  DE  to  OQ,  EF  to  OR,  .  .  .  The  figure 
ABCD  ...  is  called  the  funicular  polygon  of  the  given  forces 
with  respect  to  the  pole  O.  The  position  of  the  pole  O  is  arbi- 
trary, and  therefore  an  infinite  number  of  funicular  polygons 
may  be  drawn  with  different  poles. 

Also  the  position  of  the  point  B  in  the  line  of  action  of  Pl 
is  arbitrary,  and  hence  an  infinite  number  of  funicular  polygons 
with  their  corresponding  sides  parallel,  i.e.,  an  infinite  number 
of  similar  funicular  polygons,  may  be  drawn  with  the  same  pole. 


8 


THEORY  OF  STRUCJ^URES. 


5.  To  show  that  the  Intersection  of  the  First  and  Last 
Sides  of  the  Funicular  Polygon  (i.e.,  the  Point  G)  is  a  Point 
on  the  Actual  Resultant  of  the  System  of  Forces  Pt,  P,, 

P3,  .  .  . — First   consider  two   points  Pt,  P3,  MNP  being  the 
force  and  ABCD  the  funicular  polygon. 

Let  AB,  DC,  the  first  and  last  sides  of  the  latter,  be  pro- 


FIG. 12. 


duced  to  meet  in  gl\  also  let  DC  produced  meet  the  line  of 
action  of  Pl  in  H. 

Produce  OP  and  MN  to  meet  in  K. 

Let  the  lines  of  action  of  Pt  and  Pa  meet  in  L. 

By  similar  triangles, 


KP    _HC     KN  _  HB_      KO 

KN  ~  Hi ;    KO  ~~~~  HC  ;    KM  ~  HB 


KPKNKO  __  HC  HB  Hg, 


Hence 


or 

KP 
KM^~  HL> 

and  therefore,  since  the  angle  H  is  equal  to  the  angle  K,  the 
line  PM \s  parallel  to  the  line  Lg^. 


FRAMES  LOADED  A  T   THE  JOINTS. 


But  PM  represents  in  magnitude  the  resultant  of  the  forces 
Plt  Py,  and  is  parallel  to  it  in  direction. 

Therefore  Lg^  is  also  parallel  to  the  direction  of  the  re- 
sultant. 

But  L  is  evidently  a  point  on  the  actual  resultant  of  Pl ,  P^ . 
Hence  ^  must  be  a  point  on  this  resultant. 

Next,  let  there  be  three  forces,  P, ,  Pt ,  Pa . 

Replace  Pl ,  Pa  by  their  resultant  X  acting  in  the  direction 
Lgr  The  force  and  funicular  polygons  for  the  forces  X  and 
PI  are  evidently  MPQ  and  AgfiE,  respectively;  ana^,  the 
point  of  intersection  of  Agl  and  ED  produced,  is,  as  already 
proved,  a  point  on  the  actual  resultant  of  X  and  P9 ,  i.e.,  of 
PltP9,  and  P3 . 

Hence  the  first  and  last  sides,  AB,  ED,  of  the  funicular 
polygon  ABODE  of  the  forces  P^,  P^,  PI,  with  respect  to  the 
pole  O,  intersect  in  a  point  which  is  on  the  actual  resultant  of 
the  given  forces. 

The  proof  may  be  similarly  extended  to  four,  five,  and  any 
number  of  forces. 

If  the  forces  are  all  parallel,  the  force  polygon  of  the  two 
forces  Pl ,  P9  becomes  a  straight  line,  MNQ.  Draw  the  funicular 

M 
? 


FIG.  13. 


polygon  ABCD  as  before,  and  through  gv ,  the  intersection  of  the 
first  and  last  sides,  draw^-,  Y  parallel  to  MQ,  and  cutting  BC  in  Y. 
By  similar  triangles, 


ON 


MN 
ON 


BY 


and 


P,         QN 


ON       ON    :  CY  ' 


'  '  P. 


CY 
BY' 


10 


THEORY  OF  STRUCTURES. 


Hence  Yglt  which  is  parallel  to  the  direction  of  the  forces  Pt9 
Py,  divides  the  distance  between  their  lines  of  action  into  seg- 
ments which  are  inversely  proportional  to  the  forces,  and  must 
therefore  be  the  line  of  action  of  their  resultant.  The  proof 
may  be  extended  to  any  number  of  forces,  as  in  the  preceding. 

Funicular  Curve. — Let  the  weights  upon  a  beam  AB  become 
infinite  in  number,  and  let  the  distances  between  the  weights 
diminish  indefinitely. 

The  load  then  becomes  continuous,  and  the  funicular  poly- 
gon is  a  curve,  called  the  funicular  curve. 

The  equation  to  this  curve  may  be  found  as  follows : 

Let  the  tangents  at  two  consecutive  points  Pand  Q  meet 
in  R.  This  point  is  on  the  vertical  through  the  centre  of 
gravity  of  the  load  upon  the  portion  MN  of  the  beam. 


FIG.  14.  FIG.  15. 

Let  in  be  the  line  of  loads,  and  let  OS,  OT  be  the  radial 
lines  from  O,  the  pole,  parallel  to  the  tangents  at  P  and  Q. 
Take  A  as  the  origin. 

Let  6  be  the  inclination  of  the  tangent  at  P  to  the  beam, 
and  let  the  polar  distance  OV =  p. 

wdx  =  the  load  upon  the  portion  MN.     Then 

wdx  =  ST=  SV—  TV  —  p  tan  0  —  p  tan  (0  -f  dff) 

=  —  pdti,  approximately. 

d» 


since     u  = 


Integrating  twice, 


:,  and  £a  being  constants  of  integration. 


CENTRES  OF  GRAVITY. 


II 


If  the  intensity,  w,  of  the  load  is  constant, 


wx 


and  the  curve  is  a  parabola. 

6.  Centres  of  Gravity.  —  Let  it  be  required  to  determine 
the  centre  of  gravity  of  any  plane  area  symmetrical  with  re- 
spect to  an  axis  XX.  Divide  the  area  into  suitable  elementary 
areas  Alt  A9t  At,  .  .  .  having  known  centres  of  gravity. 


T 

FIG.  16. 


FIG.  17. 


Draw  the  force  (the  line  in)  and  funicular  polygons  corre- 
sponding to  these  areas,  and  let  g  be  the  point  in  which  the  first 
and  last  sides  of  the  funicular  polygon  meet.  The  line  drawn 
through  g  parallel  to  in  must  pass  through  the  centre  of  gravity 
of  all  the  elementary  areas  and,  therefore,  of  the  whole  area. 
Hence  it  is  the  point  G  in  which  this  line  intersects  the  axis  XX. 

Rail  and  similar  sections  may  be  divided  into  elementary 
areas  by  drawing  a  number  of  parallel  lines  at  right  angles  to 
the  axis  of  symmetry,  and  at  such  distances 
apart  that  each  elementary  figure  may,  with- 
out sensible  error,  be  considered  a  rectangle 
of  an  area  equal  to  the  product  of  its  breadth 
by  its  mean  height. 

In  the  case  of  a  very  irregular  section,  an 
accurate  template  of  the  section  may  be  cut 
out  of  cardboard  or  thin  metal.  If  the  tem- 
plate is  then  suspended  from  a  pin  through  a  point  near  the 


FIG.  1 8. 


12  THEORY  OF  STRUCTURES. 

edge,  the  centre  of  gravity  of  the  section  will  lie  in  the  vertical 
through  the  pin.  By  changing  the  point  of  suspension,  a  new 
line  in  which  the  centre  of  gravity  lies  may  be  found.  The 
intersection  of  the  two  lines  must,  therefore,  be  the  centre  of 
gravity  required.  Another  method  of  finding  the  centre  of 
gravity  is  to  carefully  balance  the  template  upon  a  needle-point. 

The  area  of  such  a  section  may  be  determined  either  by 
means  of  a  planimeter  or  by  balancing  the  template  against  a 
rectangle  cut  out  of  the  same  material,  the  area  of  the  rectangle 
being  evidently  the  same  as  that  of  the  section. 

7.  Moment  of  Inertia  of  a  Plane  Area.  —  Let  any  two 
consecutive  sides,  C£^  ,  C3674,  of  the  funicular  polygon  meet 
line  gG  in  the  points  m3,  ns. 

Let  ^  ,  .ra  ,  ;r3  ,  .  .  .  be  the  lengths  of  the  perpendiculars  from 
the  centres  of  gravity  of  Av,  At,  Az  ,  .  .  .  ,  respectively,  upon  gG. 

Draw  the  line  OH  perpendicular  to  the  line  of  loads,  and 
let  OH=p. 

By  the  similar  triangles  C9mznz  and  6^34, 


34 
—    --,-.  -,     or     »,«,=;-; 

<73  X*  X^ 

.-.  —  —  =  »wj  =  area  of  triangle  Cjnji,  . 

But  the  total  area  A  bounded  by  the  funicular  polygon 
i  .  .  .  and  the  lines  gClt  gk  is  the  sum  of  all  the  triangular 
areas  Clgml1  C9mji9,  C4mjit,  .  .  .,  described  in  the  same  manner 
as  C3m9n3. 

5*'       *,*' 
~~  p    2   "H7T" 

The  sum  2(ax*)  is  the  moment  of  inertia,  7,  of  the  plane 
area  with  respect  to  gG.     Hence, 


MOMENT  OF  INERTIA.  13 

The  moment  of  inertia  Iy  of  the  area,  with  respect  to  a 
parallel  axis  at  distance  ^  from  gG,  is  given  by  the  equation 


where  S=A1  +  A9  +  ... 

Let  the  new  axis  intersect  C^g  and  kg  in  the  points  q  and  r. 
The  triangles  qgr  and  Oin  are  similar. 

qr       in       S 

•'•^  =  7~  =  /; 

and,  therefore,  the  area  A'  of  the  triangle  qgr 


2l  2p 

Hence 


T  72          y4 

Note. — If  p  be  made  =  —  =  — , 


f=A*    and 
and 


The  angle  lOn  is  also  evidently  a  right  angle. 

8.  Cranes.  —  (a).  Jib-crane.  —  Fig.  19  is  a  skeleton  diagram  of 
an  ordinary  jib-crane.  OA  is  the  post  fixed  in  the  ground  at 
O  ;  OB  is  the  jib  ;  AB  is  the  tie.  The  jib,  tie,  and  gearing  are 
suspended  from  the  top  of  the  post  by  a  cross-head,  which 
admits  of  a  free  rotation  round  the  axis  of  the  post. 

Let  the  crane  lift  a  weight  W. 


THEORY  OF  STRUCTURES. 


Three  forces  in  equilibrium  meet  at  B ;  viz.,  W,  the  tension 
T  in  the  tie,  and  the  thrust  C  along  the  jib. 


m 
FIG.  3o. 

Draw  the  reciprocal  figure  SS&  of  B,  5,52  representing  W. 
T       SSn       AB 


and 


C_      SS,  _  BO 
AO' 


The  load  is  not  suspended  directly  from  B,  but  is  carried  by 
a  chain  passing  over  pulleys  to  a  chain-barrel  usually  fixed  to 
the  crane-post.  The  stress  5  in  the  chain  depends  upon  the 

w 

system  of  pulleys,  and  is,  e.g.,  —  ,  if  n  is  the  number  of  falls  of 

chain  from  B  and  if  friction  is  neglected.  In  order  to  obtain 
the  true  values  of  T  and  C  this  tension  5  must  be  compounded 
with  W. 

Draw  S^k  parallel  to  the  direction  in  which  the  chain  passes 
from  B  to  the  chain-barrel,  and  take  S^k  to  represent  5  in 


CRANES.  1  5 

magnitude.  The  line  Sfi  evidently  represents  the  resultant 
force  at  B  due  to  W  and  5. 

Draw  kt  parallel  to  AB. 

The  tension  in  the  tie  and  the  thrust  in  the  jib  are  now 
evidently  represented  by  tk,  tS^  ,  respectively. 

Generally  the  effect  of  chain-tension  is  to  diminish  the  ten- 
sion of  the  tie  and  to  increase  the  thrust  on  the  jib. 

The  vertical  component  of  T,  viz.,  T-^-g  —  W-^Q,  is  trans- 

mitted through  the  post. 

The  total  resultant  pressure  along  the  post  at  O 


--  =  W. 


The  pull  upon  the  tie  tends  to   upset  the   crane,  and   its 
moment  with  respect  to  O  is 


=  WAD  =  WOFy 


OF  being  the  horizontal  projection  of  AB. 

OF  is  often  called  the  radius  or  throw  of  the  crane. 

If  the  post  revolves  about  its  axis  (as  in  ///-cranes),  the  jib 
and  gearing  are  bolted  to  it,  and  the  whole  turns  on  a  pivot  at 
the  toe  G.  In  this  case,  the  frame,  as  a  whole,  is  kept  in 
equilibrium  by  the  weight  W,  the  horizontal  reaction  H  of  the 
web-plate  at  O,  and  the  reaction  R  at  G.  The  first  two  forces 
meet  in  F  and,  therefore,  the  reaction  at  G  must  also  pass 
through  F. 

Hence,  since  OFG  may  be  taken  to  represent  the  triangle 
of  forces, 

and     R= 


In  a  portable  crane  the  tendency  to  upset  is  counteracted 
by  means  of  a  weight  Q  placed  upon  a  horizontal  platform 
OL  attached  to  the  post  and  supported  by  the  tie  AL. 

The  horizontal  projection  tm  of  tk  represents  the  horizontal 


i6 


THEORY  OF  STRUCTURES. 


FIG.  21. 


pull  at  A,  and  if  tn  be  drawn  parallel  to  AL,  the  intercept  ww 
cut  off  on  the  vertical  through  m  by  the  lines  tm  and  tn  repre- 
sents the  counter-weight  required  at  L. 

(U)  Derrick- crane. — The  figure  shows  a  combination  of  a  der- 

B  rick  and  crane,  called  a  derrick- 
crane.  It  is  distinguished  from 
the  jib-crane  by  having  two 
back-stays,  AD,  AE.  One  end 
of  the  jib  is  hinged  at  or  near 
the  foot  of  the  post,  and  the 
other  is  held  by  a  chain  which 
passes  over  pulleys  to  a  winch 
on  the  post,  so  that  the  jib  may 
be  raised  or  lowered  as  required. 
The  derrick-crane  is  gener- 
ally of  wood,  is  simple  in  con- 
struction, is  easily  erected,  has 
a  vertical  as  well  as  a  lateral 
motion,  and  a  range  equal  to  a  circle  of  from  10  to  60  feet 
radius.  It  is  therefore  useful  for  temporary  works,  setting 
masonry,  etc. 

The  stresses  in  the  jib  and  tie  are  calculated  as  in  the  jib- 
crane,  and  those  in  the  back-stays  and  post  maybe  obtained  as 
follows : 

Let  the  plane  of  the  tie  and  jib  intersect  the  plane  DAE  of 
the  two  back-stays  in  the  line  AF,  and  suppose  the  back-stays 
replaced  by  a  single  tie  AF.  Take  OF  to  represent  the  hori- 
zontal pull  at  A.  -The  pull  on  the  "  imaginary"  stay  AF  is  then 
represented  by  AF  and  is  evidently  the  resultant  pull  on  the 
two  back-stays.  Completing  the  parallelogram  FGAH,  AH 
will  represent  the  pull  on  the  back-stay  AF,  and  AG  that  upon 
AD,  their  horizontal  components  being  OK,  OL,  respectively. 
The  figure  OKFL  is  also  a  parallelogram. 

If  the  back-stays  lie  in  planes  at  right  angles  to  each  other, 

OL  =  OF  cos  8  =  T  sin  a  cos  6,  and  is  a  max.  when  6  =  o°, 
and 

OK  =  OF  sin  0  =  T  sin  a  sin  6,  and  is  a  max.  when  6  =  90°, 


BRIDGE  AND  ROOF   TRUSSES.  I/ 

6  being  the  angle  FOL,  and  a  the  inclination  of  the  tie  to  the 
vertical. 

Hence  the  stress  in  a  back-stay  is  a  maximum  when  the" 
plane  of  the  back-stay  and  post  coincides  with  that  of  the  jib 
and  tie. 

Again,  let  ft  be  the  inclination  of  the  back-stays  to  the  ver- 
tical. The  vertical  components  of  the  back-stay  stresses  are 

Ts'm  a  cos  0  cot  ft     and     T sin  a  sin  6  cot  ft; 
and,  therefore,  the  corresponding  stress  along  the  post  is 
T  sin  a  cot  ft  (cos  0  -\-  sin  0), 

which  is  a  maximum  when  0  =  45°. 

9.  Shear  Legs  (or  Shears)  and  Tripods  (or  Gins)  are 


FIG.  22. 

often  employed  when  heavy  weights  are  to  be  lifted.  The 
former  consists  of  two  struts,  AD,  AE,  united  at  A  and  sup- 
ported by  a  tie  A  C,  which  may  be  made  adjustable  so  as  to 
admit  of  being  lengthened  or  shortened.  The  weight  is  sus- 
pended from  A,  and  the  legs  are  capable  of  revolving  around 
DE  as  an  axis.  Let  the  plane  of  the  tie  and  weight  intersect 
the  plane  of  the  legs  in  AF,  and  suppose  the  two  legs  replaced 
by  a  single  strut  AF.  The  thrust  along  AF.  can  now  be 
easily  obtained,  and  hence  its  components  along  the  two  legs. 

In  tripods  one  of  the  three  legs  is  usually  longer  than  the 
others.  They  are  united  at  the  top,  to  which  point  the  tackle 
is  also  attached. 

10.  Bridge  and  Roof  Trusses  of  Small  Span. — A  single 
girder  is  the  simplest  kind  of  bridge,  but  is  only  suitable  for 


IS 


THEORY  OF  STRUCTURES. 


very  short  spans.  When  the  spans  are  wider,  the  centre  of  the 
girder  may  be  supported  'by  struts  OC,  OD,  through  which  a 
portion  of  the  weight  is  transmitted  to  the  abutments. 


FIG. 


23. 


FIG.  24. 


Take  the  vertical  line  5,5,,  to  represent  P,  the  weight  at  O. 
Draw  55t  parallel  to  OC,  and  SS^  to  OD. 

Draw  the  horizontal  SH,  and  let  the  angle  AOC '=  a. 


The  thrust  along  OC  =  S^  =  S^ff  cosec  a  =  —  cosec  a. 


The  tension  along  OA  =  SH  =  S^H  cot  a  =  —  cot  of. 

The  horizontal  and  vertical  thrusts  upon  the  masonry  at  C 

P  P 

(or  D)  are  —  cot  a  and  — ,  respectively. 

If  the  girder  is  uniformly  loaded,  P  is  one  half  of  the  whole 
load. 

II.  In  the  figure  a  straining  rill,  EF,  is  introduced,  and  the 
girder  is  supported  at  two  intermediate  points. 

A  B 


FIG.  a$. 


FIG.  26. 


Let  Pbe  the  weight  at  each  of  the  points  E  and  F. 

Draw  the  reciprocal  SS^oi  the  point  E,  ^//representing  P. 


BRIDGE  AND   ROOF   TRUSSES. 


=  P.  and  the 


The  thrust  in  EC  (or  FD)  =  55,  = 


horizontal  thrust  in  the  straining  piece  =  SH  =  P =  P— 


S,H      'AC 

If  a  load  is  uniformly  distributed  over  AB,  it  may  be 
assumed  that  each  strut  carries  one  half  of  the  load  upon  AF 
(or  BE),  and  that  each  abutment  carries  one  half  of  the  load 
upon  AE  (or  BF). 

By  means  of  straining  cills  the  girders  may  be  supported  at 
several  points,  I,  2,  .  .  .  ,  and  the 
weight  concentrated  at  each  may 
be  assumed  to  be  one  half  of  the 
load  between  the  two  adjacent 
points  of  support.  The  calcula- 
tions for  the  stresses  in  the  struts, 
etc.,  are  made  precisely  as  above. 

If  the  struts  are  very  long  they  are  liable  to  bend,  and 
counterbraces,  AM,  BN,  are  added  to  counteract  this  tendency. 

12.  The  triangle  is  the  only  geometrical  figure  of  which 
the  form  cannot  be  changed  without  varying  the  lengths  of  the 
sides.  For  this  reason,  all  compound  trusses  for  bridges,  roofs, 
etc.,  are  made  up  of  triangular  frames. 

Fig.  28  represents  the  simplest  form  of  roof-truss.  AC, 
BC  are  rafters  of  equal  length  inclined  to  the  horizontal  at  an 
angle  a>  and  each  carries  a  uniformly  distributed  load  W. 


FIG.  27. 


Y 
W 


FIG.  28. 


The  rafters  react  horizontally  upon  each  other  at  C,  and 
their  feet  are  kept  in  position  by  the  tie-beam  AB.  Consider 
the  rafter  A  C. 

The  resultant  of  the  load  upon  AC,  i.e.,  W,  acts  through 
the  middle  point  D. 


2O  THEORY  OF  STRUCTURES. 

Let  it  meet  the  horizontal  thrust  //  of  BC  upon  AC  in  F. 
For  equilibrium,  the  resultant  thrust  at  A  must  also  act 
through  F. 

The  sides  of  the  triangle  AFE  evidently  represent  the 
three  forces.  Hence 

AE      WAE      W 

-  =  —  —-  =  — cot  a\ 
2  DE       2 


f  =  w^J-. 


The  thrust  R  produces  a  tension  H  in  the  tie-beam,  and  a 
vertical  pressure  W  upon  the  support. 
Also,  if  y  1S  tne  angle  FAE, 

EF         DE 


If  the  rafters  AC,  BC  are  unequal,  let  <*,,  <*a  be  their  in- 
clinations to  A,  B,  respectively. 

Let  W,  be  the  uniformly  distributed  load  upon  AC,  W9 
that  upon  BC. 


Let  the  direction  of  the  mutual  thrust  P  at  C  make  an 
angle  ft  with  the  vertical,  so  that  if  CO  is  drawn  perpendicular 


ROOF   TRUSSES. 


21 


to  FC,  the  angle  COB  =  ft\  the  angle  ACF=yo°  —AGO 
=  90°  -  (ft  -  aj. 

Draw  AM  perpendicular  to  the  direction  of  P,  and  consider 
the  rafter  A  C.  As  before,  the  thrust  Rv  at  A,  the  resultant 
weight  W^  at  the  middle  point  of  AC,  and  the  thrust  P  at  C 
meet  in  the  point  F. 

Take  moments  about  A.     Then 


But 


and 


P.AM= 
sin  ACM  =  AC  cos  (/?  —  tfx), 

-  cos  al  . 

W7!        cos  or, 


2  cos(/?  —  or,)' 
Similarly,  by  considering  the  rafter  BCy 


^ 
P 


cos 


cos 


Hence 


Wl       cos  ar,  _          W^       cos  tf2 

2  cos(/?  —  or,)  "  2   cos(/J  +  «2) ' 


and  therefore 


**/»= 


Wl  tan  a'.,  —  ^9  tan  al  ' 


The  horizontal  thrust  of  each  rafter  =  P  sin  /?. 

The  vertical  thrust  upon  the  support  A  =  Wl  —  P  cos  ft.  • 

The  vertical  thrust  .upon  the  support  B  =  Wt  +  P  cos  /?. 

13.  King-post  Truss.  —  The  simple  triangular  truss  may 
be  modified  by    introducing  a 
king-post   CO,  which    carries  a 
portion   of  the    weight    of   the 
beam     AB,    and     transfers     it 
through  the  rafters  so  as  to  act      ^ 
upon  the  tie  in  the  form  of  a  tensile  stress. 


0 
FIG.  30. 


22 


THEORY  OF  STRUCTURES. 


Let  P  be  the  weight  borne  by  the  king-post ;  represent  it 
by  CO. 

Draw  OD  parallel  to  BC,  and  DE  parallel  to  AB. 

CE       P 

DC  — =  -cosec  a  is  the  thrust  in   CA  due  to  P.  ana 

sin  a      2 

is  of  course  equal  to  DO,  i.e.,  the  thrust  along  CB. 

P 
DE  =  CE  cot  a  =  —  cot  <*   is    the    horizontal     thrust    on 

each  rafter,  and  is  also  the  tension  in  the  tie  due  to  P. 

Let  Wbe  the  uniformly  distributed  load  upon  each  rafter. 

The  total  horizontal  thrust  upon  each  rafter  =  ( W-\-  P)—  — . 

p 

The  total  vertical  pressure  upon  each  support  =  W  -\ . 

If  the  apex  C  is  not  vertically  over  the  centre  of  the  tie- 
beam  take  CO,  as  before,  to  represent  the  weight  /'borne  by 
r  the  king-post ;  draw  OD  parallel 

to  BC,  and  DE  parallel  to  AB. 
The   weight   P  produces   a 
thrust  CD  along  CA,  DO  along 
§    CB,  and  a  horizontal  thrust  DE 
upon  each  rafter. 


0 


FIG.  31. 


CE  is  the  portion  of  P  supported  at  A,  and  EO  that  sup- 
ported at  B. 

DE,  and  therefore  the  tension  in  the  tie  AB,  diminishes 

with  AO,  being  zero  when  AC 
is  vertical. 

Sometimes  it  is  expedient 
to  support  the  centre  of  the  tie- 
beam  upon  a  column  or  wall, 
the  king-post  being  a  pillar 
against  which  the  heads  of  the 
rafters  rest. 

Consider  the  rafter  AC. 
The  normal    reaction  R'  of 
CO   upon    AC,    the     resultant 
FIG.  32.  weight  W  at  the  middle  point 

D,  and  the  thrust  R  at  A  meet  in  the  point  K 


ROOF    TRUSSES. 

Take  moments  about  A.     Then 


W 


R'AC  =  W.  AE,     or     R'  =  —  cos 


Thus  the  total  thrust  transmitted  through  CO  to  the  sup- 

W 
port  at  O  is  2  —  cos  a  .  cos  or  —  £Fcos2  #. 

The  horizontal  thrust  upon  each  rafter 


W 

=  —  cos 
2 


W 
sin  a  =  —  sin  2ct. 

4 


14.  If  the  rafters  are  inconveniently  long,  or  if  they  are  in 
danger  of  bending  or  breaking  transversely,  the  centres  may 
be  supported  by  struts  ODt  OE.  A  portion  of  the  weight  upon 


0 

FIG.  33- 


FIG. 


the  rafters  is  then  transmitted  through  the  struts  to  the 
vertical  tie  (king-post  or  rod)  CO,  which  again  transmits  it 
through  the  rafters  to  act  partly  as  a  vertical  pressure  upon 
the  supports,  and  partly  as  a  tension  on  the  tie-beam.  The 
main  duty,  indeed,  of  struts  and  ties  is  to  transform  transverse 
into  longitudinal  stresses. 

This  king-post  truss  is  the  simplest  and  most  economical 
frame  for  spans  of  less  than  thirty  feet.  In  larger  spans  two 
or  more  suspenders  may  be  introduced,  or  the  truss  otherwise 
modified. 


24  THEORY  OF  STRUCTURES. 

Let  there  be  a  load  2W  uniformly  distributed  over  the 
rafters  AC,  BC>  and  assume  it  to  be  concentrated  at  the  joints 

W    W    W    W    W 
A,  D,  C,  E,  B,  in  the  proportion  —  ,  —  ,  —  ,  —  ,  -  —  . 

42224 

Also,  let  the  load  (including  a  portion  of  the  weight  upon 
the  tie-beam  AB,  and  the  weights  of  the  members  OD,  OE, 

OC)  borne  directly  at  O  be  P. 

p 
The  total  reaction  at  each  support  is  W-\-  —  -,  and  acts  in 

W 
an  opposite  direction  to  the  weight  —  there  concentrated. 

4 


Hence  the  resultant  reaction  at  a  support  is 
Thus,  the  weights  at  the  points  of  support  A  and  B  are 
taken  up  by  the  abutments,  and  need  not  be  considered  in  de- 
termining the  stresses  in  the  several  members  of  the  frame. 

Draw  the  reciprocal  SflS^  of  A.    Then 


^W     P 

= 1 — ;    ^5.,  =  tension  in  AO\ 

4        2 

5a51  =  compression  in  AD. 
Draw  the  reciprocal  5,5,535,  of  D.    Then 

5,5,  =  compression  in  OD\    5354  =  compression  in  DC; 


., 
\  -  545,  =  —  =  weight  at  D. 


Draw  the  reciprocal  5453555854  of  C.   Then 


W 
=  tension  in  CO  =  —  +  />;   5»58  =  compression  in  CE ; 

2 


W 
5854  =  —  =  weight  at  C. 


ROOF    TRUSSES. 


Draw  the  reciprocal  58S5565758  of  E.    Then 


,56  =  compression  in  OE  ;    S6»S7  =  compression  in  BE  ; 

"758  =  —  =  weight  at  E. 

» 

Draw  S6K  horizontally.    Then 


,S6  is  evidently  the  reciprocal  of  B  ;  KS,  = 
being  uie  reaction  at  B,  and  56ATthe  tension  in  the  tie  BO.    The 
reciprocal  of  O  is  also  the  figure  S^HKS^S^S^  ,  and  HK  =  P. 

15.  Collar-beams  (DE),  queen-posts  (DF,  £G),  braces,  etc., 
may  be  employed  to  prevent  the  deflection  of  the  rafters. 
The  complexity  of  the  truss  necessarily  increases  with  the  span 
and  with  the  weight  to  be  borne. 


FIG.  35, 


FIG.  36. 


With  a  single  collar-beam  and  a  uniformly  distributed  load, 
StffSs  is  the  reciprocal  of  A,  and  SlS^S3StSl  the  reciprocal  of 
D ;  S^H  being  the  reaction  at  A,  and  6^5,  the  weight  at  D. 


FIG.  37. 

With  a  collar-beam  DE,  two  king-posts  DF,  EG,  and   a 
uniformly  distributed  load,  the  stresses  at  the  joints  D  and  E 


26 


THEORY  OF  STRUCTURES. 


become  indeterminate.  To  render  them  determinate  it  is 
sometimes  assumed  that  the  components  of  the  weights  at  D 
and  E,  normal  to  the  rafters,  are  taken  up  by  the  collar-beam 
and  corresponding  king-post.  Thus  S^HS^  is  the  reciprocal  of 
A,  and  S1S^S3S4S&  the  reciprocal  of  D,  Sfl  being  the  reaction  at 
A,  SlSt  the  weight  at  D\  S4S2  is  the  normal  component  of  the 
weight,  and  the  components  of  S4S2,  viz.,  StS3  horizontal  and 
S3>S2  vertical,  represent  the  stresses  borne  by  DE  and  DF,  re- 
spectively. 

This  frame  belongs  to  the  incomplete  (Art.  1 8)  class,  and  if 
it  has  to  support  an  unequally  distributed  load,  braces  must  be 
introduced  from  D  to  G  and  from  E  to  F. 

16.  The  truss  ABC,  Fig.  40,  having  the  rafters  supported 
at  two  intermediate  points,  may  be  employed  for  spans  of  from 
30  to  50  feet.  Suppose  that  these  intermediate  points  of  sup- 
port trisect  the  rafters,  and  let  each  rafter  carry  a  uniformly  dis- 
tributed load  W. 


FIG.  39.  FIG.  40. 

Then  a  weight  may  be  considered  as  concentrated  at  each 
of  the  joints  H,  D,  C,  E,  K.     This  weight  =  — . 

Let  P  be  the  weight  directly  supported  at  each   of  the 
joints  F,  G. 

The  resultant  reaction  at  A  =  *W-}-P. 

^  is  the  reciprocal  of  A,  6V/ representing  %W  +  P. 
,  is  the  reciprocal  of  H. 
,SZ  is  the  reciprocal  of  F,  HK  representing  P,  the 

weight  directly  borne  at  F. 

S3S656S7S4S3  is  the  reciprocal  of  D,  5754  representing  the 
weight  at  D,  S<S,  the  thrust  along  HD,  S3S5  the 
tension  in  DF,  S,S,  the  thrust  along  ED,  and 
5C57  the  thrust  alon-  CD. 


INCOMPLETE  FRAMES.  2/ 

As  in  the  preceding  case,  this  truss  will  be  found  incomplete 
if  the  load  is  unevenly  distributed,  and  the  reciprocals  of  D  and 
E  will  not  close.  In  practice,  however,  the  friction  at  the  joints, 
the  stiffness  of  the  several  members,  and  the  mode  of  construc- 
tion render  the  truss  sufficiently  strong  to  meet  the  ordinary 
variations  of  load. 

17.  General  Remarks. — In  the  trusses  described  in  Arts. 
13  and  14  the  vertical  members  are  ties,  i.e.,  are  in  tension,  and 
the  inclined  members  are  struts,  i.e.,  are  in  compression.  By 
inverting  the  respective  figures  another  type  of  truss  is  obtained 
in  which  the  verticals  are  struts  while  the  inclined  members  are 
ties.  Both  systems  are  widely  used,  and  the  method  of  calcu- 
lating the  stresses  is  precisely  the  same  in  each. 

In  designing  any  particular  member,  allowance  must  be 
made  for  every  kind  of  stress  to  which  it  may  be  subjected. 
The  collar-beam  DE,  for  example,  must  be  treated  as  a  pillar 
subjected  to  a  thrust  in  the  direction  of  its  length  at  each  end  ; 
if  it  carry  a  transverse  load,  its  strength  as  a  beam,  supported 
at  the  points  D  and  E,  must  also  be  determined.  Similarly, 
the  rafters  AC,  BC,  etc.,  must  be  designed  to  carry  transverse 
loads  and  to  act  as  pillars.  But  it  must  be  remembered  that 
struts  and  queen-posts  provide  additional  points  of  support 
over  which  the  rafters  are  continuous,  and  it  is  practically  suf- 
ficient to  assume  thai  the  rafters  are  divided  into  a  number  of 
short  lengths,  each  of  which  carries  one  half  of  the  load  between 
the  two  adjacent  supports. 

When  a  tie-beam  is  so  long  as  to  require  to  be  spliced, 
allowance  must  be  made  for  the  weakening  effect  of  the  splice. 
18.  Incomplete  Frames. — The  frames  discussed  in  the 
preceding  articles  (excepting  those  referred  to  in  Art.  15)  will 
support,  without  change  of  form,  any  load  consistent  with 
strength,  and  the  stresses  in  the  several  members  can  be  found 
in  terms  of  the  load.  It  sometimes  happens,  however,  that  a 
frame  is  incomplete,  so  that  it  tends  to  change  form  under  every 
distribution  of  load.  An  example  of  this  class  is  the  simple 
trapezoidal  truss,  consisting  of  the  two  horizontal  members  AB, 
DE,  and  the  two  equal  inclined  members  AD,  BE,  Fig.  41. 
First,  let  there  be  a  weight  W  at  each  of 'the  points  D,  E. 


THEORY  OF  STRUCTURES. 


The  triangles  of  forces  for  the  joints  D  and  E,  viz.,  SSfl  and 
fl,  can  be  drawn,  and  hence  it  follows  that  there  must  be 


r 


FIG.  41. 


FIG. 


equilibrium.     This  is  also  evident  from  the  symmetrical  char- 
acter of  the  loading. 

The  same  triangles  represent  the  forces  at  the  points  of 
support  A,  B. 


.'.  reaction  at  A  =  5,11=  W  '=  S,ff  =  reaction  at  B. 

Next,  let  there  be  a  weight  Wl  at  D  and  a  weight 
<<  W,)  at  E. 


FIG.  43. 


FIG. 


It  will  now  be  found  that  the  diagram  of  forces  will  not 
close,  so  that  there  cannot  be  equilibrium.  The  joint  D  will  be 
pushed  in  and  the  frame  distorted.  The  distortion  may  be 
prevented  by  introducing  a  brace  from  A  to  E  or  from  -B  to  D. 
In  the  latter  case  S.mSS^S,  represents  the  stress-diagram,  the 
triangle  S^ffS  being  the  reciprocal  of  the  joint  E,  and  the  quad- 
rilateral 5w5,//that  of  the  joint  D.  Drawing  the  horizontal 
mn,  the  triangle  mnS,  and  the  quadrilateral  mSSji  are  evidently 
the  reciprocals  of  A  and  B,  respectively. 

.-.  nSl  =  reaction  at  A     and     nS^  =  reaction  at  B. 


INCOMPLETE  FRAMES. 


In  practice  the  loads  are  usually  transmitted  to  D  and  E  by 
means  of  two  vertical  queen-posts  (queen-rods  or  queens]  DF,  EG. 


If  there  are  no  diagonal  braces  DG,  EF,  the  distortion  of 
the  frame  under  an  unevenly  distributed  load  can  only  be  pre- 
vented by  the  friction  at  the  joints,  the  stiffness  of  the  mem- 
bers, and  by  the  queens  being  rigidly  fixed  to  AB  at  F  and  G. 

Let  Wl  be  the  load  at  F  transmitted  through  the  queen 
FD  to  D. 

Let  W^  (<  W^  be  the  load  at  G  transmitted  through  the 
queen  GE  to  E. 

If  the  frame  is  rigid,  the  reactions  Rl  at  A  and  Rt  at  B, 
which  will  balance  these  weights,  can  easily  be  found  by  taking 
moments  about  B  and  A,  successively.  Thus, 


and 


w\ 

2 


W 


W, 

2 


IV 


where  AB  =  I  and  FG  =  c. 

Draw  the  triangle  of  forces  SHS^  for  the  joint  A,  SH  rep- 
resenting R^ . 

The  triangle  SS^X  is  the  reciprocal  of  the  joint  at  /?,  and  the 
tension  in  FD  should,  therefore,  be  XS,  —  SH=  R, .  But  the 


30  THEORY  OF  STRUCTURES 

tension  in  FD  is  actually  Wl ,  so  that  there  is  an  unbalanced 
force, 

JW  _  iv     I  —  c 

-  wi  —  Ri  =        2         ~7~' 

acting  along  FD. 

To  take  up  this  unbalanced  force  and  render  the  frame  rigid 
the  diagonal  DG  is  introduced,  and  the  stress  for  which  it 
should  be  designed  is  evidently 

W  --  W  I  —  c  9 
(W,  -  *,)  sec  FDG  =  -  '  „',  ^, 


s  being  the  length  of  the  diagonal  and  d  the  depth  of  the  truss. 
The  complete  stress  diagram  is  as  shown  in  Fig.  46. 

Cor.  i.  The  manner  in  which  distortion  is  prevented  by  the 
stiffness  of  AB  may  be  shown  as  follows: 

Let  x  be  the  force  of  resistance  which  AB,  by  its  stiffness, 
can  exert  at  F  or  G  against  any  load  which  tends  to  make  it 
deviate  from  the  horizontal. 

If  Wis  the  load  at  Ft  the  actual  downward  pull  upon  D  is 
W—  x  ;  this  must  necessarily  produce  an  equal  upward  pull  at 
E,  which  must  be  balanced  by  the  force  of  resistance  x  at  G, 


and 

W 

'    2' 

Thus  the  beam  AB  will  be  acted   upon  by  an  upward  pull 

W 

—  at  F  and  an  equal  downward  pull  at  G,  forming  a  couple 

W 
of  moment  —  c,  and  showing  that  equilibrium  is  impossible. 

The  upward  reaction  R^  at  A  is 

l(-^+f       WJ^\-    j^£ 

1  ~~    /  V  2    ~  2        ~  ~2    '   1     I   ~~2~l 


COMPOSITE  FRAMES. 

-—-  downward  reaction  at  B,  and  the  moment  at  F  (or  G) 
Wcl-c       We.. 


Cor.  2.  Let  a  weight  Wbe  supported  at  the  joint  D  of  any 
quadrilateral  frame  ADEB.     Draw  the  reciprocal  SS^  of  D, 


w 


sv 


FIG.  47. 


FIG.  48. 


S^S^  representing  W.  Draw  553  parallel  to  EB  and  intersecting 
the  vertical  5t5,  produced  in  ,S3 .  The  weight  which  can  be 
borne  at  E  consistent  with  equilibrium  is  represented  by  S^S3 . 

19.  Composite  Frames  or  Trusses  (i.e.,  frames  made  up 
of  two  or  more  simple  frames). — An  example  of  this  class  has 
already  been  given  in  the  case  of  the  king-post  roof  (Art.  13). 

Bent  Crane. — Fig.  49  shows  a  convenient  form  of  crane 
when  much  head-room  is  required  near  the  post.  The  crane 
is  merely  a  semi-girder,  and  may  be  tubular  with  plate-webs  if 
the  loads  are  heavy,  or  its  flanges  may  be  braced  together  as 
in  the  figure  for  loads  of  less  than  ten  tons.  The  flanges  may 
be  kept  at  the  same  distance  apart  throughout,  or  the  distance 
may  be  gradually  diminished  from  the  base  towards  the  peak. 

Let  the  numbers  in  Fig.  50  denote  the  stresses  '  \  the  cor- 
responding members.  Three  forces,  Slt  C9,  and  W,  act  through 
the  point  (i),  so  that  Sl  and  £72  may  be  obtained  in  terms  of 
W;  three  forces,  5,,  52,  T3 ,  act  through  (2),  so  that  S2  and  Ts 
may  be  obtained  in  terms  of  Sj  and  therefore  of  W\  four 
forces,  S2,  £T2,  53,  £74 ,  act  through  (3),  and  the  values  of  52 ,  Ct 


32 


THEORY  OF  STRUCTURES. 


being  known,  those  of  S9,  C4  may  be  determined.  Proceed- 
ing in  this  way,  it  is  found  that  of  the  forces  at  each  succeed- 
ing joint  only  two  are  unknown,  and  the  values  of  these  are 
consequently  determinate. 


FIG.  50. 

The  calculations  may  be  checked  by  the  method  of  moments. 
and  by  the  stress  diagram  (Fig.  50). 
E.g.,  let  W—  10  tons. 
Take  moments  about  the  point  (7).    Then 


or  T,  =    -     =  26  tons  =  (68)  in  Fig.  50. 


No  other  forces  enter  into  the  equation  of  moments,  as  the 
portion  of  the  crane  above  a  plane  intersecting  (68)  and  passing 
through  (7)  is  kept  in  equilibrium  by  the  weight  of  10  tons 
and  the  stresses  T7,  56,  C6  ;  the  moments  of  S6  and  C6  about 
(7)  are  evidently  zero. 

In  the  stress  diagram  (Fig.  50)  PaQ  is  the  reciprocal  of  the 
point  i,  abQ  of  the  point  2,  PcbQ  of  3,  Qbcd  of  4,  and  so  on. 

Other  examples  of  composite  roof  and  bridge  frames  will 
now  be  given. 

20.  Roof-trusses.  —  A  roof  consists  of  a  covering  and  of 
trusses  (or  frames)  by  which  it  is  supported.  The  covering 
is  generally  laid  upon  a  number  of  common  rafters  which  rest 


ROOF   TRUSSES. 


33 


upon  horizontal  beams  (or  purlins),  the  latter  being  carried 
by  trusses  spaced  at  intervals  varying  with  the  type  of  con- 
struction but  averaging  about  10  ft.  The  truss  rafters  are 
called  principal  rafters,  and  the  trusses  themselves  are  often 
designated  as  principals. 

In  roofs  of  small  span  the  trusses  and  purlins  are  sometimes 
dispensed  with. 

Types  of  Truss. — A  roof-truss  may  be  constructed  of  tim- 
ber, of  iron  or  steel,  or  of  these  materials  combined.  Timber 
is  almost  invariably  employed  for  small  spans,  but  in  the 
longer  spans  it  has  been  largely  superseded  by  iron,  in  con- 
sequence of  the  combined  lightness,  strength,  and  durability 
of  the  latter. 

Attempts  have  been  made  to  classify  roofs  according  to  the 
mode  of  construction,  but  the  variety  of  form  is  so  great  as  to 
render  it  impracticable  to  make  any  further  distinction  than 
that  which  may  be  drawn  between  those  in  which  the  reac- 
tions of  the  supports  are  vertical  and  those  in  which  they  are 
inclined. 


FIG.  51. 


FIG.  52. 


FIG.  53. 


FIG.  57. 


FIG.  58. 


FIG. 


59- 


Fig.  51  is  a  simple  form  of  truss  for  spans  of  less  that  30  ft. 

Fig.  52  is  a  superior  framing  for  spans  of  from  30  to  40  ft.; 
it  may  be  still  further  strengthened  by  the  introduction  of 
struts,  Figs.  53  and  54,  and  with  such  modification  has  been 
employed  to  span  openings  of  90  ft.  It  is  safer,  however,  to 
limit  the  use  of  the  type  shown  by  Fig.  53  to  spans  of  less  than 


34  THEORY  OF  STRUCTURES. 

60  ft.  Figs.  55,  56,  57,  58,  and  59  are  forms  of  truss  suitable 
for  spans  of  from  60  to  100  ft.  and  upwards. 

Arched  roofs,  Figs.  58  and  59,  admit  of  a  great  variety  of 
treatrrent.  They  have  a  pleasing  appearance,  and  cover  wide 
spans  without  intermediate  supports.  The  flatness  of  the 
arch  is  limited  by  the  requirement  of  a  minimum  thrust  at  the 
abutments.  The  thrust  may  be  resisted  either  by  thickening 
the  abutments  or  by  introducing  a  tie.  If  the  only  load  upon 
a  roof-truss  were  its  own  weight,  an  arch  in  the  form  of  an 
inverted  catenary,  with  a  shallow  rib,  might  be  used.  But  the 
action  of  the  wind  induces  oblique  and  transverse  stresses,  so 
that  a  considerable  depth  of  rib  is  generally  needed.  If  the 
depth  exceed  12  in.,  it  is  better  to  connect  the  two  flanges  by 
braces  than  by  a  solid  web.  Roofs  of  wide  span  are  occasion- 
ally carried  by  ordinary  lattice-girders, 

Principals,  Purlins,  etc. — The  principal  rafters  in  Figs.  51 
to  57  are  straight,  abut  against  each  other  at  the  peak,  and  are 
prevented  by  tie-rods  from  spreading  at  the  heels.  When 
made  of  iron,  tee  (T),  rail,  and  channel  (both  single  i — i  and 
double  ][)  bars,  bulb-tee  (T)  and  rolled  (I)  iron  beams,  are  all 
excellent  forms. 

Timber  rafters  are  rectangular  in  section,  and  for  the  sake 
of  economy  and  appearance,  are  often  made  to  taper  uniformly 
from  heel  to  peak. 

The  heel  is  fitted  into  a  suitable  cast-iron  skew-back,  or  is 
fixed  between  wrought-iron  angle-brackets  (Figs.  60,  61,  62), 
and  rests  either  directly  upon  the  wall  or  upon  a  wall-plate. 


FIG.  60.  FIG.  61.  FIG.  62. 

When  the  span  exceeds  60  ft.,  allowance  should  be  made 
for  alterations  of  length  due  to  changes  of  temperature.     This 


ROOF    TRUSSES. 


35 


may  be  effected  by  interposing  a  set  of  rollers  between  the 
skew-back  and  wall-plate  at  one  heel,  or  by  fixing  one  heel  to 
the  wall  and  allowing  the  opposite  skew-back  to  slide  freely 
over  a  wall-plate. 

The  junction  at  the  peak  is  made  by  means  of  a  casting  or 
wrought-iron  plates  (Figs.  63,  64,  65). 


FIG.  63. 


FIG.  64. 


FIG.  65. 


Light  iron  and  timber  beams  as  well  as  angle-irons  are  em-, 
ployed  as  purlins.     They  are  fixed  to  the  top  or  sides  of  the 
rafters   by  brackets,  or   lie   between   them  in  cast-iron  shoes 
(Figs.  66  to  71),  and  are  usually  held  in  place  by  rows  of  tie- 


FIG.  70.  FIG.  71.  FIG.  72. 

rods,  spaced  at  6  or  8  ft.  intervals  between  peak   and  heel, 
running  the  whole  length  of  the  roof. 

The  sheathing  boards  and  final  metal  or  slate  covering  are 
fastened  upon  the  purlins.  The  nature  of  the  covering  regu- 
lates the  spacing  of  the  purlins,  and  the  size  of  the  purlins  is 
governed  by  the  distance  between  the  main  rafters,  which  may 


THEORY  OF  STRUCTURES. 


vary  from  4  ft.  to  upwards  of  25  ft.  But  when  the  interval 
between  the  rafters  is  so  great  as  to  cause  an  undue  deflection 
of  the  purlins,  the  latter  should  be  trussed.  Each  purlin  may 
fee  trussed,  or  a  light  beam  may  be  placed  midway  between  the 
main  rafters  so  as  to  form  a  supplementary  rafter,  and  trussed 
-as  in  Fig.  72. 

Struts  are  made  of  timber  or  iron.  Timber  struts  are 
rectangular  in  section.  Wrought-iron  struts  may  consist  of 
J_-irons,  T-bars,  or  light  columns,  while  cast-iron  may  be  em- 
ployed for  work  of  a  more  ornamental  character.  The  strut- 
heads  are  attached  to  the  rafters  by  means  of  cast  caps, 
wrought-iron  straps,  brackets,  etc.  (Figs.  73  to  76),  and  the 
strut-feet  are  easily  designed  both  for  pin  and  screw  Conner 
tions  (Figs.  77  to  80). 


FIG.  73- 


FIG.  74. 


FIG.  75. 


FIG.  76. 


FIG.  77. 


FIG.  78. 


FIG. 


FIG.  80. 


Ties  may  be  of  flat  or  round  bars  attached  either  by  eyes 
and  pins  or  by  screw  ends,  and  occasionally  by  rivets.  The 
greatest  care  is  necessary  in  properly  proportioning  the  dimen- 
sions of  the  eyes  and  pins  to  the  stresses  that  come  upon  them. 

To  obtain  greater  security,  each  of  the  end  panels  of  a  roof 
snay  be  provided  with  lateral  braces,  and  wind-ties  are  often 
made  to  run  the  whole  length  of  the  structure  through  the 
feet  of  the  main  struts. 


ROOF    WEIGHTS.  37 

Due  allowance  must  be  made  in  all  cases  for  changes  of 
temperature. 

21.  Roof-weights. — In  calculating  the  stresses  in  the 
different  members  of  a  roof-truss  two  kinds  of  load  have  to  be 
dealt  with,  the  one  permanent  and  the  other  accidental.  The 
permanent  load  consists  of  the  covering,  \he  framing,  and  ac- 
cumulations of  snow. 

Tables  at  the  end  of  the  chapter  show  the  weights  of  various 
coverings  and  framings. 

The  weight  of  freshly  fallen  snow  may  vary  from  5  to  20 
Ibs.  per  cubic  foot.  English  and  European  engineers  consider 
an  allowance  of  6  Ibs.  per  square  foot  sufficient  for  snow, 
but  in  cold  climates,  similar  to  that  of  North  America,  it  is 
probably  unsafe  to  estimate  this  weight  at  less  than  12  Ibs. 
per  square  foot. 

The  accidental  or  live  load  upon  a  roof  is  the  wind-pressure, 
the  maximum  force  of  which  has  been  estimated  to  vary  from 
40  to  50  Ibs.  per  square  foot  of  surface  perpendicular  to  the 
direction  of  blow.  Ordinary  gales  blow  with  a  force  of  from  20 
to  25  Ibs.,  which  may  sometimes  rise  to  34  or  35  Ibs.,  and  even 
to  upwards  of  50  Ibs.  during  storms  of  great  severity.  Press- 
ures much  greater  than  50  Ibs.  have  been  recorded,  but  they 
are  wholly  untrustworthy.  Up  to  the  present  time,  indeed,  all 
wind-pressure  data  are  most  unreliable,  and  to  this  fact  may  be 
attributed  the  frequent  wide  divergence  of  opinion  as  to  the 
necessary  wind  allowance  in  any  particular  case.  The  great 
differences  that  exist  in  all  recorded  wind-pressures  are  pri- 
marily due  to  the  unphilosophic,  unscientific,  and  unpractical 
character  of  the  anemometers  which  give  no  correct  informa- 
tion either  as  to  pressure  or  velocity.  The  inertia  of  the  mov- 
ing parts,  the  transformation  of  velocities  into  pressures,  and 
the  injudicious  placing  of  the  anemometer,  which  renders  it 
subject  to  local  currents,  all  tend  to  vitiate  the  results. 

It  would  be  practically  absurd  to  base  calculations  upon 
the  violence  of  a  wind-gust,  a  tornado,  or  other  similar  phe- 
nomena, as  it  is  almost  absolutely  certain  that  a  structure 
would  not  lie  within  its  range.  In  fact,  it  may  be  assumed 
that  a  wind-pressure  of  40  Ibs.  per  square  foot  upon  a  surface 


38  THEORY  OF  STRUCTURES. 

perpendicular  to  the  direction  of  blow  is  an  ample  and  perfectly 
safe  allowance,  especially  when  it  is  remembered  that  a  greater 
pressure  than  this  would  cause  the  overthrow  of  nearly  all  the 
existing  towers,  chimneys,  etc. 

22.  Wind-pressure  upon  Inclined  Surfaces.  —  The  press- 
ure upon  an  inclined  surface  may  be  obtained  from  the  follow- 
ing formula,  which  was  experimentally  deduced  by  Hutton, 
viz.  : 


pn  =p  sin 


p  being  the  intensity  of  the  wind-pressure  in  pounds  per  square 
foot  upon  a  surface  perpendicular  to  the  direction  of  blow,  and 
pn  being  the  normal  intensity  upon,  a  surface  inclined  at  an 
angle  a  to  the  direction  of  blow. 

Let/A,/T  be  the  components  of  /„,  parallel  and  perpen- 
dicular, respectively,  to  the  direction  of  blow^ 

•  '.  PH  =  pn  sin  a,     and    p,  —  pn  cos  a. 

« 

Hence,  if  the  inclined  surface  is  a  roof,  and  if  the  wind  blows 
horizontally,  a  is  the  roof's  pitch. 

Again,  let  v  be  the  velocity  of  a  fluid  current  in  feet  per 
second,  and  be  that  due  to  a  head  of  h  feet. 

Let  w  be  the  weight  of  the  fluid  in  pounds  per  cubic  foot. 

Let  p  be  the  pressure  of  the  current  in  pounds  per  square 
foot  upon  a  surface  perpendicular  to  its  direction. 

If  the  fluid,  after  striking  the  surface,  is  free  to  escape  at 
right  angles  to  its  original  direction, 


p  =  2/lW  =  —  W. 

Hence  for  ordinary  atmospheric  air,  since  w  =  .08  lb.,  approx- 
imately, 


-•°§,._f.!'v 

'  32      ~UoJ 


DISTRIBUTION  OF  LOADS.  39 

When  the  wind  impinges    upon    a  surface  oblique  to  its 

Iv    sin  /?\2 
direction,  the  intensity  of  the  pressure  is  ^ — — — j  ,  v  being 

the  absolute  impinging  velocity,  and  ft  being  the  angle  between 
the  direction  of  blow  and  the  surface  impinged  upon.  (See 
chapter  on  Bridges.) 

Tables  prepared  from  formulae  A  and  B  are  given  at  the 
end  of  the  chapter. 

23.  Distribution  of  Loads. — Engineers  have  been  accus- 
tomed to  assume  that  the  accidental  load  is  uniformly  dis- 
tributed over  the  whole  of  the  roof,  and  that  it  varies  from  30 
to  35  Ibs.  per  square  foot  of  covered  surface  for  short  spans, 
and  from  35  to  40  Ibs.  for  spans  of  more  than  60  ft.  But  the 
wind  may  blow  on  one  side  only,  and  although  its  direction  is 
usually  horizontal,  it  may  occasionally  be  inclined  at  a  con- 
siderable angle,  and  be  even  normal  to  a  roof  of  high  pitch. 
It  is  therefore  evident  that  the  horizontal  component  (/A)  of 
the  normal  pressure  (pn)  should  not  be  neglected,  and  it  may 
cause  a  complete  reversal  of  stress  in  members  of  the  truss, 
especially  if  it  is  of  the  arched  or  braced  type. 

If  PH  is  the  total  normal  wind-pressure  on  the  side  of  a 
roof  of  pitch  a,  its  horizontal  component  Pn  sin  a  will  tend  to 
push  the  roof  horizontally  over  its  supports.  This  tendency 
must  be  resisted  by  the  reactions  at  the  supports. 

In  roofs  of  small  span,  the  foot  of  each  rafter  is  usually  fixed 
to  its  support,  and  it  may  be  assumed  that  each  support  exerts 

the  same  reaction,  which  should  therefore  be  equal  to  -  ! • 

In  roofs  of  large  span  the  foot  of  one  rafter  is  fixed,  while  that 
of  the  other  rests  upon  rollers.  The  latter  is  not  suited  to  with- 
stand a  horizontal  force,  and  the  whole  of  the  horizontal  com- 
ponent of  the  wind-pressure  must  be  borne  at  the  fixed  end, 
where  the  reaction  should  be  assumed  to  be  equal  to  Pn  sin  a. 

In  designing  a  roof-truss  it  is  assumed  that  the  wind  blows 
on  one  side  only,  and  that  the  total  load  is  concentrated  at  the 
joints  (or  points  of  support)  of  the  principal  rafters. 

E.g.,  let  the  rafters  AB,  AC  of  a  truss  be  each  supported  at 


40  THEORY  OF  STRUCTURES. 

two  intermediate  points  (or  joints),  D,  E  and  F,  G,  respectively, 
and  let  the  wind  blow  on  the  side  AB. 


FIG.  81. 


=  FG  =  l^  EA  =  GA  =/,;  and 
let  /1  +  /3  +  /8  =  /;  •'•  BC  =  2!  cos  a,  a  being  the  angle 
ABC. 

Let  W  be  the  permanent  (or  dead)  load  per  square  foot  of 
roof-surface. 

Let  pn  be  the  normal  wind-pressure  per  square  foot  of  roof- 
surface. 

Let  d  be  the  horizontal  distance  in  feet  from  centre  to 
centre  of  trusses. 

The  total  normal  live  load  concentrated 


2 

at  E  =  pnd— ;  at  A  = 

j     The  total  vertical  dead  load  concentrated  at  D  and  F  = 


--    ;  at  E  and  G  =  wd±- ;  at  A  =  wdl,. 


Let  R^ ,  R^  be  the  resultant  vertical  reactions  at  B  and  £T, 
respectively  (i.e.,  the   total  vertical  'reactions   less   the  dead 

weights  \wd-)  concentrated  at  these  points). 


DISTRIBUTION  OF  LOADS.  41 

Take  moments  about  C. 

-.  R^.1  cos  a  =  sum  of  moments  of  live  loads  about  C-\-  sum 

of  moments  of  dead  loads  about  C, 
=  moment  of  resultant  wind-pressure  about  C 
-f-  moment  of  resultant  dead  load  about  (7, 


=  PJd  ~  +  /  cos  2a    -f  wd(l,  +  2/,  +  2/8)/  cos  a, 

where  —  +  /  cos  2a  is  the  perpendicular  from  C  upon  the  line 

of   action   of   the    resultant  wind-pressure   which   bisects  AB 
normally. 

(N.B.  The  moment  of   the  horizontal  reaction  at  B  or  C 
about  C  is  evidently  nil.) 

R^  may  be  found  by  taking  moments  about  B. 

To  determine  the  stresses  in  the  various  members  of  a  roof- 
truss  two  methods  may  be  pursued  : 

(x)  A  single  stress  diagram  may  be  drawn  to  represent  the 
combined  effect  of  the  live  and  dead  loads.  This  will  be  found 
to  be  the  quickest  and  most  useful  method. 

(y)  The  normal  wind-pressure  (/„)  may  be  resolved  into  its 
vertical  (pv)  and  horizontal  (/A)  components  ;  pv  may  then  be 
combined  with  the  dead  load  W,  and  a  stress  diagram  drawn 
for  the  vertical  loads  only.  A  second  diagram  may  be  drawn 
for  the  horizontal  loads.  The  resultant  stresses  will  be  the 
algebraic  sum  of  the  corresponding  stresses  in  the  two  dia- 
grams. 

A  third  method  will  be  referred  to  in  a  subsequent  article. 

24.  Ex.  I.  Method  (x)  applied  to  the  roof-truss  ABC, 
Fig.  82. 

The  dead  load  =  wld  concentrated  at  A. 

The  live  loads  —  pn—  acting  at  each  of  the  points  A  and 

B,  normally  to  AB. 

The  vertical  reaction  at  B 


wld        pnld  /i       cos  2d\ 

--  --    --  ---  /  • 

2         '     COS  <*\  2       / 


THEORY  OF  STRUCTURES. 


Let  rollers  be  placed  underneath  C. 

Tne  total  horizontal  reaction  =  pnld  sin  a,  and  is  wholly 
borne  at  B. 


FIG.  82. 


FIG.  83. 

At  B  there  z.rz  five  forces  in  equilibrium,  of  which  three  are 
known,  and  the  reciprocal  of  B  may  be  thus  described  : 

Draw   SjS.,   to   represent  the   normal  wind-pressure  \pn—\ 

at  B  ^  5258  to  represent  Rl  ;  58S4  to  represent  the  horizontal 
reaction  (pnld  sin  a) ;  S4S6  parallel  to  BD ;  5355  parallel  to 
AB. 

The  closed  figure  5,5,5,5^5,  is  the  reciprocal  required,  and 
the  stresses  in  BD,  AB,  at  B,  are  represented  by  5453,  S}S5, 
respectively,  being  a  tension  and  a  thrust. 

At  D  there  are  three  forces  in  equilibrium,  of  which  the 
tension  in  DB  has  been  found.  Drawing  5456  horizontally 
and  S5S6  parallel  to  AD,  the  triangle  545855  is  evidently  the 
reciprocal  of  D,  the  stresses  in  DA,  DE  being  represented  by 
5556,  5654,  respectively,  and  being  both  tensions. 


ROOF-TRUSSES.  43 

Again,  the  triangle  StS6S,  is  the  reciprocal  of  E,  the  stresses 
in  EC,  EA  being  represented  by  5457  ,  5657  ,  respectively,  and 
being  both  tensions. 

At  A  there  are  six  forces  in  equilibrium,  of  which  two,  viz., 

the   normal   pressure,  \pn  —  J,  and   the  dead  weight,  (wld\  are 
given,  while  the  stresses  in  AB,  AD,  AE  have  been  found. 


Draw  595i  to  represent  pn  —  ,  and  S6S9  to  represent  wld. 

Five  of  the  forces  at  A  are  therefore  represented  by  the 
following  lines,  taken  in  order  :  S8Sd,  SstS1  ,  5j55,  SbS6  ,  SCS7  . 

Hence  the  closing  line  S,SS  must  necessarily  represent  in 
direction  and  magnitude  the  force  in  AC  at  A,  and  it  is  a 
thrust. 

Also,  StS6S^  must  be  the  reciprocal  of  C,  and  therefore  StSfs 
represents  the  reaction  at  C. 

The  resultant  reaction  at  B  is  represented  in  direction  and 
magnitude  by  5Q54  . 

The  line  5859  must  pass  through  the  point  S4  ,  as  S3S4  ,  the 
horizontal  reaction,  is  merely  the  horizontal  projection  of  5952  , 
the  total  wind-pressure. 

The  dotted  lines  show  the  altered  stresses  if  rollers  are 
under  B,  the  end  C  being  fixed.  The  stress  in  each  member  is 
diminished,  and  as  the  truss  should  be  designed  to  meet  the 
most  unfavorable  case,  the  stresses  should  be  calculated  on  the 
assumption  that  the  rollers  are  on  the  leeward  side. 

This  may  be  considered  an  invariable  rule  for  roof-trusses. 

Ex.  2.  Method  (x)  applied  to  the  roof-truss  ABC,  Fig.  84. 

wld 
The    vertical    dead    load  —  -   at    each    of    the    points 

F,  A,  G. 

The   live  load,  acting  normally  to  AB,  =  —  —  at  each  of 

4 

the  points  B  and  A,  and  =  -&—  at  F. 
The  vertical  reaction  R^  at  B 

=  lwld  +      Pnld      (COS  2<X  +  l). 

2  cos  a 


44 


THEORY  OF   STRUCTURES. 


The  horizontal  reaction  at  B  =  pnld  sin  a,  rollers  being 
under  C  as  before. 


FIG.  85. 

Describe  the  stress  diagram  in  precisely  the  same  manner 
as  in  Ex.  i. 

Taking  S,S2  to  represent  the  normal  wind-pressure  at  B, 
S^S3  "  "          "    vertical  reaction  R^  at  B, 

5354  "  "          "    horizontal  reaction  at  B, 

/.  5I5a53545651        is  the  reciprocal  of  By 
c  c  c  c  c  c  «  «  "7^ 

OiOsOp.OgO,  F, 

c  c  e  c  c  «  «          «   n 

^5w->4^9w56°5  J^1 

?       C       C         «  «  "       >4 

>no12o7  A} 

c          «  <*          «  /c 


c  e  c    c    c 

O9|J4W->13V->10°9 

C    C      C 

O4O14013 


"  £, 
"  C. 


S,St  is  the  horizontal  projection  of  5,5,  +  S,S,  +  5,5,., ,  i.e., 
of  the  total  normal  wind-pressure,  and  therefore  the  vertical 
through  5la  must  pass  through  5, . 


ROOF   TRUSSES. 


45 


The  dotted  lines  show  the  altered  stresses  if  rollers  are 
under  B. 

The  resultant  reaction  at  B  is  represented  in  direction  and 
magnitude  by  SzSt. 

Ex.  3.  Method  (x)  applied  to  the  truss  represented  by 
Fig.  86. 


FIG.  86. 


Data.  —Pitch  =  30° ;  AD  =  BD  —  AE  —  CE  =  23  ft. ; 
trusses  13  ft.,  centre  to  centre  ;  dead  weight  =  8  Ibs.  per  square 
foot  of  roof-surface  ;  wind-pressure  on  one  side  of  roof  (say  AB] 
normal  to  roof-surface  =  28  Ibs.  per  square  foot;  DF-=-DH 
=  EG  =  EK\  DF  and  EG  are  vertical  ;  rollers  under  one  end, 
say  C\  span  =  79  ft. ;  AF=BH  —  21  ft,  nearly  ;  FH  =  3$  ft., 
nearly. 

Total    live   load  =  4459  Ibs. ^  =  -       13.28]  at    each    of   the 
points  F,  H, 

and  =  3822  lbs.(  =  —  .  13  .  28]    at    each    of    the 
points  A,  B. 

Total  dead  load  =  1274  lbs.(  =  ~  .  13  .  8J   at    each    of    the 
points  F,  H,  K,  G, 

and  =  2184  Ibs.  (  =  21  .  13  .  8)  at  the  point  A. 


THEORY  OF  STRUCTURES. 


ROOF   TRUSSES.  47 

Resultant  vertical  reaction  at  B 

=  £(4  X  1274  +  2184)  +  —  |TT^  =  13201.8  Ibs. 

Horizontal  reaction  at  B  —  16562  sin  30°  =  8281  Ibs. 
Let  i  inch  represent  16,000  Ibs.,  and  on  this  scale  draw 


S^S^  =  3822  Ibs.,  the  normal  wind-pressure  at  B  ; 
5253  =  13201.8  Ibs.,  the  vertical  reaction  at  B  ; 
5354  =  8281  Ibs.,  the  horizontal  reaction  at  B  \ 
S4S6  parallel  to  BD,    and    v&  parallel  to  BA. 


The  figure  S^S.S.S.S,  is  the  reciprocal  of  JB. 
The  stress  diagram  can  now  be  easily  completed,  the  recip 
rocals  of  the  points  H,  F,  D,  A,  G,  K,  £,  and  C  being 


516514513517516  ,    S17S18S19S16S17  ,     and    S19SJ8S4S17  ,    respectively. 

St  ,  as  before,  is  in  the  vertical  line  S19S16  produced. 
On  the  assumed  scale, 

S4S5  =  the  tension  at  BD  ;  S,Sb  =  thrust  in  BH; 

5M5W=    "        "         "  AD-,          5658=      "       "  HF\ 
=    "         "         "  BE',          S9Sn=      "       "   AF; 

S6S0=      "       «  DH-, 


These  are  the  maximum  stresses  to  which  the  members  of 
one  half  of  the  truss  can  be  subjected,  and  for  which  they 
should  be  designed.  It  is  also  usual,  except  in  special  cases,  to 
make  the  two  halves  symmetrical. 

5254  is  the  resultant  reaction  at  B. 

If  the  end  C  is  fixed  and  rollers  placed  under  B,  the  reduced 
stresses  may  be  shown  by  dotted  lines  as  in  Exs.  i  and  2. 


48 


THEORY  OF  STRUCTURES. 


Exs.  4  and  5.  Method  (x)  applied  to  the  trusses  repre- 
sented by  Figs.  88  and  90. 

It  is  assumed,  as  before,  that  there  is  a  normal  wind-press- 
ure upon  AB,  and  that  rollers  are  under  C. 

Figs.  89  and  91  are  the  maximum  stress  diagrams  corre- 
sponding to  Figs.  88  and  90,  respectively,  and  are  drawn  in  pre- 
cisely the  same  manner  as  described  in  the  preceding  examples. 

Remark  on  Fig.  88. — The  stresses  at  the  joints  F  and  D 
are  indeterminate,  and  it  is  assumed  that  the  stress  in  FL 


Si 


FIG.  89. 


is  equal  to  that  in  FH.  The  reciprocal  of  F  thus  becomes 
^io^ii^6^8^9^n^i3^io>  Si*Si»(=  ^8^9)  being  the  stress  in  FD.  This 
truss  is  an  example  of  a  frame  with  redundant  bars,  in  which 
the  stresses  can  only  be  determined  when  the  relative  yield  of 
the  bars  is  known. 


ROOF   TRUSSES. 


49 


Remark  on  Fig.  90. — The  stress-diagram,  Fig.  91,  for  each  of 
the  joints  in  the  horizontal  BC  (Fig.  90)  is  closed  by  the  return 
of  one  side  upon  another.  Thus  at  D  the  stress  diagram  is 
,  the  closing  line  5459  (the  tension  in  DE)  returning 


FIG.  91. 

upon    55vS4  (the   tension   in  DB).     The  total  stress  in  AE  is 

evidently  represented    by  512513,   the    reciprocal   of  A    being 
c   c   c   c    c    c 

°14015010012013°14' 

Ex.  6.  A  truss  with  curved  upper  and  lower  chords,  the 
portions,  however,  between  consecutive  joints  being  assumed 
straight. 

Under  a  uniformly  distributed  load  the  truss  (Fig.  92)  is 
evidently  incomplete,  and  the  stress  diagrams  at  the  joints  in 
the  lower  chord  will  not  close,  so  that  equilibrium  is  impossible. 
The  frame  is  made  complete  and  the  stresses  determinate  by 
introducing  ties  as  in  Fig.  93,  the  corresponding  stress  diagram 
for  one  half  the  truss  being  shown  by  Fig.  94. 

Next,  let  there  be  a  wind-pressure  on  the  side  AB  of  the 
truss.  In  order  to  prevent  a  reversal  of  stress  in  the  diagonal 
ties  on  the  side  AC  (Fig.  93),  additional  ties  DE,  FG,  called 
counter -braces,  are  introduced  as  in  Fig.  95.  Fig.  96  gives  the. 


THEORY   OF  STRUCTURES, 


stress  diagram   due   to   wind-pressure   only,   it   being  assumed 
that  the  end  C  rests  upon  rollers  and  that  B  is  fixed. 


FIG.  92. 


FIG.  93. 


FIG.  94. 

Note. — 21  —  wind-press,  at  B  = 

23  =          "  "M  = 

34=  "M  = 

45  = 
56  = 
67  —  "  "  A  = 


"  O  = 
11  O  = 


FIG.  95. 


wind-press,  upon  BM, 
"  BM, 
"  MO, 

"  "  MO, 

"  OA, 

"  "      OA. 


\K  —  vertical  reaction,  at  B, 
HK  •=.  horizontal  reaction  at  B. 

Ex.  7.  A  single  example  will  serve  to  illustrate  method  (y). 
Take  the  truss  represented  by  Fig.  97. 

Fig.  98  is  the  stress  diagram  due  to  the  vertical  load  upon 
the  roof,  viz.,  the  dead  weight  -{-  vertical  component  of  wind- 
pressure.  pq  is  the  vertical  reaction  at  B  and  is 


ff 


2AH)d. 


ROOF-  TR  USSES. 

qm  is  the  weight  at  F  and  is 


=  weight  at  H  = 


Fig.  99  is  the  stress-diagram  due  to  horizontal  component 
of  wind-pressure,  rollers  being  placed  under  B  and  the  end  C 
being  fixed. 


p'o'  =  downward  reaction  at  B  = 


pnld  sin2  a 
4    cos  OL 


o'q'  •=.  horizontal  force  of  wind  at  B  =  ^—      —BF\ 


q'm' 


=  m'n'=  horizontal  force  of  wind  at  ForH=  — -&ff. 

2 


Total  resultant  stresses  in  the  members  BF,  FH,  HA,  DF, 
DH,  DB,  JDA,  DE  are  represented  by  qr  —  q'r' ,  ms  —  m's'y 
nt  —  n't',  sr  —  s'r',  st  —  st',  pr  —  p'r',  tv  —  /V,  pv  —  p V, 
respectively. 

Note. — The  stress  diagrams  for  trusses  with  both  of  the  lower 
ends  of  the  principal  rafters  fixed,  are  drawn  in  precisely  the 
same  manner  as  described  in  the  preceding  examples. 


THEORY  OF   STRUCTURES. 


Thus,  in  Fig.  100,  S^S^S^  is  the  reciprocal  of  A, 
representing  the  portion  of  the  horizontal  wind-pressure  borne 


FIG.  100. 

at  A.  Again,  HS6S,S,H  is  the  reciprocal  of  B,  HS^  represent- 
ing the  portion  of  the  horizontal  wind-pressure  borne  at  C. 
HSZ  =  HSt  +  S4S3  =  total  horizontal  wind-pressure,  5253  repre- 
senting the  vertical  reaction  at  B,  and  HS^  that  at  C. 

25.  Bridge-trusses. — A  bridge-truss  proper  consists  of 
an  upper  chord (or  flange),  a  lower  chord  (or  flange],  and  an  in- 
termediate portion,  called  the  web,  connecting  the  two  chords. 
Its  depth  is  made  as  small  as  possible  consistent  with  economy, 
strength,  and  stiffness.  Its  purpose  is  to  carry  a  distributed 
load,  which,  as  in  the  case  of  roof-trusses,  is  assumed  to  be 
concentrated  at  the  joints,  or  panel-points,  of  the  upper  and 
lower  chord.  Trussed  beams  are  also  employed  for  the  same 
object,  and  examples  of  simple  frames  of  this  class  have  already 
been  given. 

The  following  are  bridge-trusses  of  a  more  complex  char- 
acter. 

Ex.  I.  The  beam  BC (Fig.  101)  is  supported  at  three  points 
by  the  vertical  struts  DF,  AK,  EG,  which  are  tied  at  the  feet 
by  the  rods  DB,  DK,  AB,AC,  and  EK,  EC.  Let  W,,  W,,  W, 
be  the  loads  concentrated  at  the  joints  F,  K,  G,  respectively. 
Draw  the  line  of  loads  S,S, ,  S,St  being  W, ,  S2S3  =  W^ ,  and 

C    C    _    TX7 
0304  —    VV  i  . 

Describe  the  funicular  polygon  with  any  pole  O,  and  draw 
OH  parallel  to  the  closing  line  MN  of  this  polygon.  Then 


BRIDGE    TRUSSES. 


53 


}  is  the  reaction  at  B  and  HS^  the  reaction  at  £7  (Art.  3). 
is  the  reciprocal  of  B,  S^S&  being  the  thrust  along 
FB,  and  ^/^  the  tension  along  BD. 

5  is  the  reciprocal  of  Ft  5,52  being  W19  the 
weight  at  Ft  52S6  the  thrust  along  KF,  SeS6  the 
thrust  along  DF. 

H  is  the  reciprocal  of  D,  56S7  being  the  tension 
along  DK,  and  S,H  the  tension  along  Z^^. 
is  the  reciprocal  of  A,  S,Se  being  the  thrust  along 
KA,  and  S^H  the  tension  along  AE. 

So,  S2S3S9S8S7S6S2 ,  S3S4S10S9S3 ,  S9S]0HS,S9 ,  and  S4//5IO  are 
the  reciprocals  of  K,  G,  E,  and  C,  respectively,  the  closing  line 
51054  being  necessarily  horizontal  and  representing  the  stress 
in  GC. 


FIG.  101.  FIG.  102. 

This  truss  inverted  is  often  used  for  bridge  purposes  in  dis- 
tricts where  timber  is  plentiful,  as  it  may  be  constructed 
entirely  of  wood.  The  stresses  in  the  several  members  of  the 
inverted  truss  are  of  course  reversed  in  kind  but  unchanged  in 
magnitude,  and  are  given  by  the  same  stress  diagram. 

Note. — The  reactions  ffSlt  HS^  may  be  obtained  at  once  by 
the  method  of  moments.  Thus,  by  taking  moments  about  C, 
the  reaction  R,  at  B  is 


and  by  taking  moments  about  B,  the  reaction  R^  at  C  is 


54 


THEORY  OP    STRUCTURES. 


Ex.  2.  In  the  truss  represented  in  the  accompanying  figure, 
the  length  of  the  beam  AB  is  so  great  that  the  single  triangu- 
lar truss  ACB  with  a  single  central  strut  CO  is  an  insufficient 
support.  The  two  halves  are  therefore  strengthened  by  the 
simple  triangular  trusses  AGO  with  a  central  strut  GF  and 
BPO  with  a  central  strut  PN. 

Again,  each  quarter-length,  viz.,  AF,  FO,  ON,  NB,  is  simi- 
larly trussed.  The  subdivisions  may,  if  necessary,  be  carried 
still  farther.  This  truss  in  four,  'eight,  sixteen,  .  .  .  divisions  or 

D  F.SH          0  L  N  Q  B 


panels  is  known  as  the  Fink  truss,  and  has  been  widely  em- 
ployed in  America,  the  number  of  panels  usually  being  eight 
or  sixteen. 

The  members  shown  by  the  dotted  lines  may  be  introduced 
for  stiffness,  and  the  platform  may  be  either  at  the  top  or 
bottom.  The  weight  directly  borne  by  a  strut  is  usually  de- 
termined from  the  loads  upon  the  two  adjacent  panels  by 
assuming  the  corresponding  portions  of  the  beam  to  be  inde- 
pendent beams  supported  at  the  ends.  Thus  if  there  be  a 
weight  PFat  the  point  5  in  the  panel  FH,  the  portion  of  W 
borne  by  the  strut  GF  at  F  is 

Sff 


W 


FH' 


and  the  portion  borne  by  the  strut  KH  at  H  is 

FS 


W 


FH 


Let  Wlt  W,,  W,,  W.,  Wb,  W6,  W,  be  the  weights  upon 
the  struts  (or  posts)  DE,  FG,  HK,  OC,  LM,  NP,  QR,  respect- 
ively. 

Let  PltPttPt9P4tPbfPtt  P,  be  the  compressions  to  which 
these  posts  are  severally  subjected. 


FINK    TRUSS.  55 

Let  a,  /?,  y  be  the  inclinations  to  the  vertical  of  AE,  AG, 
AC,  respectively. 

Let  Tlt  T^  T3,  .  .  .  be  the  tensions  in  the  ties,  as  in  Fig.  103. 

The  tensions  in  the  ties  meeting  at  the  foot  of  a  post  are 
evidently  equal. 

Each  triangular  truss  may  be  considered  separately. 

From  the  truss  AEF,  2T,    cos  a  =  P1  —  W,  ; 

from  the  truss  A  GO,  2  T,    cos  /?=/>,=  W,+(  T,+  T,)  cos  a  ; 

from  the  truss  FKO,  2  T3  cos  a  =  P3  =  W3\ 

from  the  truss  ACB, 


from  the  truss  OMN,     2  T,  cos  a  =  P,  =  W^', 
from  the  truss  OPB,      2T6cos/3=P6=  W6+(Tb+T,)  cos  a  ; 
from  the  truss  NRB,     2  T,  cos  a  =  P,=  W,  . 
Hence 

W 


lJw  -1-  W*+W*   I    ^ 

4"t" 


GCC  y, 


=  --  sec 


and  the  values  of  P,  ,  P2,  P,,  .  .  .  can  be  at  once  found. 


56  THEORY  OF  STRUCTURES. 

Again,  the  thrust  along  AF=  7",  sin  a-\-Tt  sin  /?  +"  jT4  sin  7  ; 
at  ,F  =  T,  sin  )3  +  T.  sin  7  ; 

along  FO=  T,  sin  /?  +  Tt  sin  y  +7;  sin  a  -, 
"  at  6>  =  r4  sin  p ; 

etc.,         etc. 

If  the  truss  carries  a  uniformly  distributed  load  W, 


W 
,  =  -     sec  a, 


;  =  T;  =  —  sec  /?,       T;  =  —  sec 
o  4 


If  the  above  diagram  is  inverted,  it  will  represent  another 
type  of  truss  in  which  the  obliques  are  struts  and  the  verticals 
ties. 

Note. — The  stresses  in  the  several  members  of  each  of  the 
trusses  due  to  the  weight  it  is  designed  to  carry,  may  of  course 
be  easily  determined  graphically  in  the  manner  already  de- 
scribed in  previous  articles. 

Ex.  3.  Fig.  104  represents  a  beam  trussed  by  a  number 
of  independent  triangular  trusses,  the  vertical  posts  being 


FIG. 


104. 


equidistant.  The  weight  concentrated  at  the  head  of  each 
post  may  be  found  by  the  method  described  in  Ex.  2,  which 
in  fact  is  generally  applicable  to  all  bridge  and  roof  trusses. 

Let  7*,,  7*a  be  the  tensions  in  AE,  BE,  respectively. 

Let  Wl  be  the  weight  at  D. 


WARREN    TRUSS. 


57 


Let  ofl ,  #2  be  the  inclinations  of  AE,  BE,  respectively,  to 
the  vertical. 


sin 


1  sin  (or,  +  or,) ' 


T  = 


D         F 
FIG.  105. 


Similarly,  the  stress  in  any  other  tie  may  be  obtained. 

The  compression  in  the  top  chord  is  the  algebraic  sum  of 
the  horizontal  components  of  all  the  stresses  in  the  ties  which 
meet  at  one  end. 

The  verticals  are  always  struts  and  the  obliques  ties. 

This  truss  has  been  used  for  bridges  of  considerable  span, 
but  the  ties  may  prove  inconveniently  long. 

Ex.  4.  The  figure  SANT  represents  an  ordinary  triangular 
truss  of  the  Warren  type,  supported  at  the  ends  5  and  T. 

Draw   the   line   of  loads    16,   12      A      c      E     .G      L       N 
being   the  weight   at  B  and  23,  34,  AAAAAA 
45,  56  the  weights  at  D,  F,  K,  My  \ 
respectively. 

With  any  pole  O  describe  the 
funicular  polygon  and  draw  OP  par- 
allel to  its  closing  line  QR. 

.'.  [Pis  the  reaction  at  5,  and  6P 
that  at  T. 

The  reciprocal  of  S  is  the  triangle 
Pi  Sl  ;  15,  being  the  tension  in  SB, 
and  5,/^the  compression  in  AS. 

The  reciprocal  of  A  is  the  triangle  Q 
PS}S^ ;  5,52  being  the  tension  in  AB, 
and  S^Pthc  compression  in  CA. 

The  reciprocal  of  B  is  the  figure 
5,i2585a5, ;  253  being  the  tension  in 
BD,  5352  the  compression  in  CB,  and 
12  the  weight  at  B.  FIG.  107. 

The  reciprocal  of  C  is  the  figure  PS,2S3S4P',  5354  being  the 
tension  in  CD,  and  Sf  the  compression  in  EC. 

The  reciprocal  of  D  is  the  figure  5323555453  ;  355  being  the 
tension  in  DF,  5554  the  compression  in  ED,  and  23  the  weight 
at  D. 


/    \ 


\  / 


V 

SQ 


FIG.  106. 


THEORY   OF  STRUCTURES. 


The  reciprocal  of  E  is  the  figure  PStSbS6P-  S6S6  being  the 
tension  in  EF,  and  S6P  the  compression  in  GE. 

The  reciprocal  of  F  is  the  figure  S^S,tS<.S.0  ;  48,  being  the 
tension  in  FK,  5,56  the  tension  in  FG,  and  34  the  weight  at  F. 
And  so  on,  the  closing  line  PSn  for  the  reciprocal  of  T  being 
necessarily  parallel  to  NT. 

The  arrow-heads  show  the  character  of  the  stresses  in  the 
several  members  of  the  truss. 

Note.  —  The  reactions  may  also  be  at  once  determined  by 
the  method  of  moments. 


Thus 
and 


iP  =  |(I2)  +  4(23)  +  1(34)  +  1(45)  + 

6P  =  4(12)  +  f(23)  +  f  (34)  +  £(45)  +  f  (56). 


(12)      (23)      (34),    (45)      (56)      (CV)      (78) 

FIG.  108. 


Ex.  5.  In  the  truss  represented  by  the  accompanying  figure, 
the  joints  in  the  upper  as  well  as 
those  in  the  lower  chord  are  loaded, 
the  weights  being  transmitted  to  the 
former  by  means  of  vertical  sus- 
penders. 

Fig.  109  is  evidently  the  corre- 
sponding stress  diagram. 

Note.  —  In  the  trusses  repre- 
sented by  Figs.  1 06  and  109,  the  floor 
is  carried  upon  the  lower  chords.  If 

the  trusses  are   inverted,  the  floor 

* 

may  be  carried  on  the  upper  chords. 
The  stresses  in  the  several  members 
are  evidently  the  same  in  magnitude 
and  are  only  reversed  in  kind. 
FlG-  I09-  Ex.  6.  The  Howe    truss  repre- 

sented by  Fig.  no  is  very  widely  used  and  maybe  constructed 
of  timber,  of  iron,  or  of  timber  and  iron  combined. 

A       C       E       G        L       N        R 


ciXAA* 


/V/6 

Qlp 


B        D      'F       H       K       M       Q 


HOWE    TRUSS. 


59 


Let  there  be  a  uniformly  distributed  load  upon  the  truss 
consisting  of  a  weight  W  at  each  of  the  joints  B,  D,  ...  in  the 
lower  chord. 

The  reaction  at  each  support  =  3^  W. 

Fig.  in  is  the  stress  diagram,  and  the  several  members  of 
the  truss  are  indicated  on  the  lines  representing  the  stresses  to 


BS 


i 
— % 


W 


W 


w. 


AC 


~CE 


W 


w 


MQ 


W 


QT 


DF 


#/  8 


FH 


KM 


FIG.  in. 

which  they  are  subjected.  The  directions  of  these  stresses  at 
the  joints,  and  hence  also  their  character,  are  easily  determined 
by  following  in  order  the  sides  of  the  reciprocals.  The  verti- 
cals are  evidently  all  ties  and  the  diagonals  all  struts. 

If  the  load  is  unevenly  distributed,  the  stresses  in  different 
members  may  be  reversed.     For  example, 


BD 


T 


DF 
FIG.  ii2. 


-s1- 


FIG.  113. 


Let  the  truss  carry  a  single  weight  P  at  any  point  D. 

The  reciprocal  of  D  is  SlS.tS^StS^Sl  (Fig.  112),  5,52  represent- 


6c 


THEORY   OF   STRUCTURES. 


ing  Pt  and  the  arrow-heads  showing  the  directions  of  the  forces 
now  acting  at  D.  Thus  the  force  in  DE  at  I),  represented  by 
,S'3S4 ,  acts  from  D  towards  E,  and  is,  therefore,  a  tension. 

Hence,  in  order  that  DE  may  not  be  subjected  to  a  tensile 
force,  counterbraces  CF,  EH  are  introduced  so  that  the  por- 
tion of  P  borne  on  the  support  at  T  may  be  transmitted 
through  the  system  CFEH  to  H  and  from  H  to  T  through  the 
rcgiilar  system  HGKLMNQRT.  The  reciprocal  of  D  is  now 
51525856  (Fig.  1 1 3),  and  the  reciprocal  of  Cthe  figure  HS,S,S,Sfl, 
the  arrow-heads  showing  the  directions  of  the  forces  at  C.  It 
will  be  at  once  observed  that  FC  must  be  a  strut. 

In  order  to  make  provision  for  a  varying  load,  as  when  a 
train  passes  over  a  bridge,  counterbraces  are  introduced  in  the 
panels  on  both  sides  of  the  centre,  and  although  they  may  not 
be  necessary  in  every  panel,  they  will  give  increased  stiffness  to 
the  truss. 

Note. — Generally  speaking,  a  panel  is  that  portion  of  the 
bridge-truss  between  two  consecutive  verticals,  and  the  ends 
of  the  verticals  are  called  panel-points. 

Ex.  7.  Fig.  114  represents  a  Pratt  truss,  and  is  merely  an 
inverted  Howe  truss.  The  diagonals  become  ties  and  the 


FIG.  114. 


verticals   struts.     Counterbraces    are  introduced  to  resist   th^ 
action  of  a  varying  load,  precisely  as  described  in  Ex.  6. 

Ex.  8.   The  bowstring  truss  in  its  simplest  form  is  repre- 


M      Q 


FIG.  115. 


sented  by  Fig.  115.     Assuming  that  the  portions  of  the  upper 
chord   between  consecutive  joints  are  straight,  the  stress  dia- 


BOWSTRING   TRUSS. 


61 


gram  for  a  uniformly  distributed  load  and  for  one  half  the 
truss  is  Fig.  116. 

The  panels,  however,  are  incomplete  frames,  and  if  the  truss 

8.  S. 

1 


J 


S8 


FIG.  116. 


FIG.  117. 


has  to  carry  an  unequally  distributed  load,  ties  similar  to  that 
shown  by  the  dotted  line  MNmust  be  introduced  in  the  several 
panels  in  order  to  prevent  distortion. 

For  example,  let  there  be  a  single  load  P  at  the  joint  N, 
and  let  there  be  no  brace  NM.  The  stress  in  the  first  vertical 
is  evidently  nil.  The  reciprocal  of  ^V  is  S1S^S3S^S5S1 ,  Fig.  117, 
S2S3  representing  P.  The  reciprocal  of  L  is  HS^S^S^H,  and  the 
arrow-heads  show  the  directions  of  the  forces  at  H. 

Thus  the  force  in  OL,  which  is  represented  by  S4S6 ,  acts  from 
O  towards  L,  and  is,  therefore,  a  compression.  But,  under  a 
uniformly  distributed  load,  the  diagonals  are  all  ties,  and  NM 
is  introduced  to  take  up  that  portion  of  P  which  would  be 
otherwise  transmitted  through  LO  in  the  form  of  a  compression. 
In  this  case  the  reciprocal  of  L  is  HS^S^H,  since  the  stress  in  LO 
due  to  P  is  assumed  to  be  nil.  Also  the  reciprocal  of  N  is 
SlS^S^S^S^S^Sl.  The  stress  in  NM,  represented  by  S8S7,  acts 
from  N  to  M  and  is  a  tension. 

Hence  the  diagonals  NM  are  also  ties,  and  the  portion  of 
the  weight  P  borne  at  L  is  carried  to  Q  through  the  system 
NMOQ. 

Ex.  9.  Fig.  118  is  a  bowstring  truss  with  isosceles  bracing. 
Under  an  arbitrary  load  Fig.  119  is  the  stress  diagram,  the 
loads  at  a,  b,  c,  d,  e,  f,  g  being  12,  23,  34,  45,  56,  67,  78,  respect- 
ively. As  in  the  Warren  girder,  the  diagonals  may,  under  the 
action  of  a  varying  load,  be  subjected  to  both  tensile  and  com- 


62 


THEORY   OF   STRUCTURES. 


pressi-ve  stresses.     They  must,  therefore,  be  designed  to  bear 
such  reversal  of  stress. 


a       .b         c        d        e        f       g 
FIG.  118. 


FIG. 


It  is  assumed,  as  before,  that  the  portions  of  the  upper 
chord  between  consecutive  joints  are  straight. 

Note.  —  The  design  of  bridge-trusses  will  be  further  con- 
sidered in  a  subsequent  chapter. 

26.  Method  of  Sections.  —  It  often  happens  that  the 
stresses  in  the  members  of  a  frame  may  be  easily  obtained  by 
the  method  of  sections.  This  method  depends  upon  the 
following  principle  : 

If  a  frame  is  divided  by  a  plane  section  into  two  parts,  and 
if  each  part  is  considered  separately,  the  stresses  in  the  bars 
(or  members)  intersected  by  the  secant  plane  must  balance  the 
external  forces  upon  the  part  in  question. 

Hence  the  algebraic  sums  of  the  horizontal  components, 
2(X),  of  the  vertical  components,  2(Y\  and  of  the  moments 
of  the  forces  with  respect  to  any  point,  2(M),  are  severally 
zero  ;  i.e.,  analytically, 

=  o,     S(Y)  =  o,     and     2(M)  =  o. 


These  equations  are  solvable,  and  the  stresses  therefore 
determinate,  if  the  secant  plane  does  not  cut  more  than  three 
members. 


EXAMPLES.  63 

Ex.  I.  ABC  is  a  roof-truss  of   60  ft.  span  and  30°  pitch. 


The  strut  DF  =  GH  =  5  ft.;  the  angle  FDA  =90°.  Also 
AF=F£  =  AG  =  GC. 

The  vertical  reaction  at  .#  =  5  tons.  The  weight  concen- 
trated at  D  —  4f  tons. 

Let  the  angle  ABF  '=  a. 

AB  =  30  sec  30°  =  20  VI  ;     cot  a  =  —  —?  =  2  1/J, 


.*.  sm  a  =      .  _  ;       cos  a  = 


. 

4/13 

If  the  portion  of  the  truss  on  the  right  of  a  secant  plane 
MNbe  removed,  the  forces  C,  T^  ,  7^  in  the  members  AD,AF, 
FG  must  balance  the  external  forces  5  tons  and  4^  tons  in  order 
that  the  equilibrium  of  the  remainder  of  the  truss  may  be  pre- 
served. 

Hence,  revolving  horizontally  and  vertically, 

7;+  T,  cos  (a  +  30°)  -  ^sin6o0=o; 
T,  sin  (a  +  30°)  —  C  cos  60°  +  5  -  4^  =  0. 
Taking  moments  about  Ft 

C.$-  $BF  cos  (30°  -  «)  +  4kDF  sin  30°  =  o. 
But 
cos  («+3o°)  =  sin  (0300)=:,  Cos  (30°  —)= 


64  THEORY   OF  STRUCTURES. 

BF  =  BD  sec  a  —  5  VT$,      and   DF  =  5  tt. 


T.. 


Hence  £7=  15^  tons,  7^  =  9.89  tons,  and  7",  =  6.35  tons. 

Ex.  2.  The  figure  represents  a  portion  of  a  bridge-truss  cut 
,     off  by  a  plane  MN  and  supported  at 


409,400  Ibs 


FiG. 


the  abutment  at  A. 

The  vertical  reaction  at  A 

=  409,400  Ibs. 

T   F          The  weight  at  B  =  49,500  Ibs. 
"  C  =  38,700  Ibs. 


=  BC=  24  ft.  ;  BD=24  ft.;  CE  =  29%  ft. 


The  forces  C',  D'  ,  T  in  the  members  met  by  MN  must 
balance  the  external  forces  at  A,  B,  C. 
Revolving  horizontally  and  vertically, 

T+D'  cos  a—  C  cos/5  =  o; 
D'  sin  a  +  C  sin  ft  —  409400  +  495°°  +38/00  =  o  ; 

a  and    /3  being    the  inclinations  to  the  horizon  of  EF,  DE, 
respectively. 

Taking  moments  about  Ey 


—  T  x  291  +  409400  x  48  —  49500  x  24  =  o. 


PIERS. 


But 


.-.  sin  a  — 


Hence 


«  =         =          and     tan  ft  =       =  -. 
24        9  24       9 


II  9  2 

,  __  ,    cos  a  —  —  ^,    sin  p  =—  —  -,   cos  p 
202  4/202  1/85 

T  =  629,427^  Ibs.  ; 


981^00 


a 

^= 

V8S 


27.  Piers.  —  To  determine  the  stresses  in  the  members  of 
the  braced  piers  (Fig.  122)  supporting  a  deck  bridge. 


40  tons 


FIG.  122. 


FIG,  123, 


Data. — Height  of  pier  =  50  ft. ;  of  truss  =  30  ft.     Width 
of  pier  at  top  —  17  ft. ;  at  bottom  =  33!  ft. 


OO  THEORY  OF  STRUCTURES. 

The  bridge  when  most  heavily  loaded  throws  a  weight  of 
IPO  tons  on  each  of  the  points  A  and  B. 

Weight  of  half-pier  =  30  tons. 

The  increased  weight  at  each  of  the  points  C,  D  and  E,  F, 
jfrom  the  portions  AD  and  CF  of  the  pier  =  5  tons. 

Resultant  horizontal  wind-pressure  on  train  =  40  tons  at 
iS/j-  feet  above  base. 

Resultant  horizontal  wind-pressure  on  truss  =  20  tons  at 
65  feet  above  base. 

Resultant  horizontal  wind-pressure  on  pier  =  2^  tons  at 
each  of  the  points  C  and  E. 

With  the  wind-pressure  acting  as  in  the  figure,  the  diagonals 
CB,  ED,  and  GF  are  required.  When  the  wind  blows  on  the 
other  side,  the  diagonals  D  to  A,  F  to  C,  and  H  to  E  are 
brought  into  play.  The  moment  of  the  couple  tending  to 
overturn  the  pier 

=  40  X  8;J  +  20  X  65  +  4  X  25  =  4900  ton-feet. 

•22.2. 

The  moment  of  stability  =  (200  -f  30)  X  ~-  =  3871!  ft.-tons. 

Thus  the  difference,  =  4900  —  3871!  =  IO28|  ft.-tons,  must 
be  provided  for  in  the  anchorage.  The  pull  on  a  vertical 


anchorage-tie  at  G  ==•  -—jf  —  3°r5oT  tons- 
33s 

Again,  if  H  be  the  horizontal  force  upon  the  pier  at  A  due 
to  wind-pressure, 

H  X  50  =  40  x  8;J  +  20  x  65  =  4800  ; 
H  =  96  tons. 

The  stress  diagram  can  now  be  easily  drawn. 

The  reciprocals  of  the  points  A,  B,  C,  D,  E,  F  are  4321, 
2561,  11-10-4169,  65789,  13-12-11-98-14,  and  87-15-16-14,  respec- 
tively. In  the  stress  diagram  43  =  96  tons,  32  =  25  =  100  tons, 
57  —  7-15=4-10—11-12  —  6  tons,  and  10-1  1  =  12-13  =  2i 
tons.  The  stress  in  EG  is  of  an  opposite  kind  to  the  stresses 
in  AC,  CE. 


WEIGHT  OF  ROOF-COVERINGS. 


67 


. — In  computing  the  stresses  in  the  leeward  posts  of  a 
braced  pier,  it  is  usual  in  American  practice  to  assume  that  the 
maximum  load  is  upon  the  bridge  and  that  the  wind  exerts  a 
pressure  of  30  Ibs.  per  sq.  ft.  upon  the  surfaces  of  the  train  and 
structure,  or  a  pressure  of  50  Ibs.  per  sq.  ft.  upon  the  surface 
of  the  structure  alone.  The  negative  stresses  in  the  windward 
posts  of  the  pier  are  determined  when  the  minimum  load  is  on 
the  bridge,  the  wind-pressure  remaining  the  same. 

TABLE   OF    WEIGHTS   OF    ROOF-COVERINGS. 


Description  of  Covering. 

Weight  of 
Covering 
in  Ibs.  per 
sq.  ft.  of 
Covered 
Area. 

Dead  Weight  of  Roof  in  Ibs.  per  sq. 
ft.  of  Covered  Area. 

Boarding  ($~inch) 

2c  to  "\ 

Boarding  and  sheet-iron  

6  5 

JC 

8  to  -1.25 

Corrugated  iron  and  laths 

5e 

Felt   asphalted       .  .        

^  tO     4. 

Felt  and  gravel   

8  to  10 

Galvanized  iron 

I  to  3 

Laths  and  plaster  ... 
Pantiles  

9  to  10 
6  to  10 

<;  to  8 

Sheet-zinc     . 

I  25  to  2 

Sheet-iron  (corrugated).      .. 

3      A 

8  without  boards  and  11  with  boards 

K                     (i 

3X 

for  spans  up  to  75  ft. 
12  without  boards  and  15  with  boards 

Sheet-iron  (16  W.G.)  and  laths. 
Shingles  (i6-inch)   .    . 

5 

2 

for  spans  from  75  to  150  ft. 
10  on  laths  for  spans  up  to  75  ft 

"       (lone') 

14    on    laths   for    spans    from    75    to 

Sheathing  (i-inch  pine).  ..?... 

a 

150  ft. 

"         (chestnut  and  maple) 
(ash,  hickory,  oak),. 
Slates  (ordinary)  

4 

5 

e  to  Q 

13  without  boards  or  on  laths  and  16 

Slates  (large)        . 

Q  to  1  1 

on  i^-in.  boards  for  spans  up  to 
.75ft- 

Slates  and  iron  laths  

IO 

on  i£-in.  boards  for  spans  from 
75  to  150  ft. 

Thatch       

6  S 

Tiles  

7  to  2O 

Tiles  and  mortar 

oe  fO  ^O 

Timbering  of    tiled    and    slate 
roofs  (additional)  

5c   Jo  6  5 

68 


THEORY   OF  STRUCTURES. 


WEIGHTS   OF   VARIOUS   ROOF-FRAMINGS. 


Description  of  Roof. 

Location. 

Cover- 
ing. 

Span. 

Width 
of 
Bays. 

Weight  in  Ibs. 
per  sq.  ft.  of 
Covered  Area. 

Pitch. 

Fram- 
ing. 

Cover- 
ing. 

Pent  

ft.    in 
15   0 
37  o 
40  o 
50  o 

53  3 

54  o 
55  o 
72  o 

62  o 

76  o 
79  o 
80  8 

90   2 

84  o 

100  0 

130  o 
50  o 

154   0 
211    0 

97  o 

153    0 

41  o 
81  6 

I2O   O 

72  o 
240  o 
45  o 

ft.    in. 

5  o 
12   0 
IO  O 

II    0 

14  o 
6  6 

20   0 
12    0 

25  o 
13  c 
ii  8 

20  o 

9  o 
14  o 
26  o 

II    O 

26  o 
24  o 
13  o 
26  o 
16  o 
24  o 

29  4 
14  6 

3-5 
4.6 

5-5 
3-0 

2.085 

9-5 
ii.  6 
7.0 

3-013 

2.6 

3.  86 
4.72 

5.00 

5.66 

7.72 
5.42 

12.  I 

30° 

30° 

30° 
30° 
26°  34' 

«             « 

«             « 

«             « 

j  Liverpool  \ 
1     Docks      f 

Felt 

«             it 

«             « 

«             « 

J       Timber     ") 
^y\    rafters  and  1 
^^^^   |    struts,  iron  f 

ties 
«                      <« 

«                      <« 
<«                      « 

Common  Truss          .  .  .  . 

j  Liverpool  ) 
1     Docks     1 

ii 
ii 

« 

Zinc 

Zinc 
Zinc 
Slates 

13.6 

3-5 
7-0 
6.4 
9.6 
4-9 

II.  O 
12.  O 
15-0 
10.7 

16.8 
ii.  8 
ii.  3 
24-5 
11-5 

>  <            i< 

«            «« 

Bowstring                    .  .  . 

Manchester 
Lime  Street 
Birmingham 
Strasburg 
Paris 
Dublin 
Derby 
Sydenham 

St.  Pancras 
Cremorne 

1,1 

K 

Arched 

ii 

« 

« 



« 

« 

« 

WIND   PRESSURES. 


69 


TABLE  OF  THE  VALUES  OF  Pn,  Pv ,  Ph,  IN  LBS.  PER  SQ.  FT,  OF 

SURFACE,   WHEN   P  -  40,  AS    DETERMINED    BY 

THE  FORMULA  Pn  =  P  .  sin  a'-84  cosa-i. 


Pitch  of  Roof. 

Pn 

Pv 

Ph 

5° 

5-0 

4.9 

•4 

10° 

9-7 

9.6 

i-7 

20° 

18.1 

17.0 

6.2 

30° 

26.4 

22.8 

13-2 

40° 

33-3 

25-5 

21.4 

50° 

38.1 

24-5 

29.2 

60° 

40.0 

20.0 

34-0 

70° 

41.0 

I4.O 

38.5 

80° 

40.4 

7.0 

39-8 

90° 

40.0 

0.0 

40.0 

TABLE   PREPARED   FROM   THE   FORMULA 


-(=)•• 


Velocities  in 
feet  per  second. 

Velocities  in 
miles  per  hour. 

Pressure  in 
Ibs.  per  sq.  ft. 

10 

6.8 

•25 

20 

13-6 

1.  00 

40 

27.2 

4.00 

60 

40.8 

g.OO 

70 

47.6 

12.25 

80 

54-4 

16.00 

90 

61.2 

20.25 

100 

68.0 

25.00 

no 

74.8 

30-25 

1  20 

81.6 

36.00 

130 

88.4 

42.25 

150 

102.0 

56.25 

JO  THEORY  OF  STRUCTURES. 


EXAMPLES. 

1.  Show  that  .the  locus  of  the  poles  of  the  funicular  polygons  of 
which  the  first  and  last  sides  pass  through  two  fixed  points  on  the  clos- 
ing line,  is  a  straight  line  parallel  to  the  closing  line. 

2.  The  first  and  last  sides  of  a  funicular  polygon  of  a  system  of  forces 
intersect  the  closing  line  in  two  fixed  points.    Show  that  for  any  position 
of  the  pole  each  side  of  the  polygon  will  pass  through  a  fixed  point  on 
the  closing  line. 

3.  Four  bars  of  equal  weight  and  length,  freely  articulated  at  the 
extremities,  form  a  square  ABCD,     The  system  rests  in  a  vertical  plane, 
the  joint  A  being  fixed,  and  the  form  of  the  square   is   preserved  by 
means  of  a  horizontal  string  connecting  the  joints  B  and  D.     If   W  be 
the  weight  of'each  bar,  show  (a)  that  the  stress  at  C  is  horizontal  and 

=  — ,  (b)  that  the  stress  on  EC  at  B  is      WjL^L  and  makes  with  the  ver- 

2  2 

tical  an  angle  tan  -J-J,  (<r)  that  the  stress  on  AB  at  B  is      W^1^  and 

makes  with  the  vertical  an   angle  tan"1!,  (d)   that   the   stress  upon 
AB  at  A  is  f  W,  (e)  that  the  tension  of  the  string  is  2  W . 

4.  Five  bars  of  equal  length  and  weight,  freely  articulated  at  the 
extremities,  form  a  regular  pentagon  ABCDE.     The  system  rests  in  a 
vertical  plane,  the  bar  CD  being  fixed  in  a  horizontal  position,  and  the 
form  of  the  pentagon  being  preserved  by  means  of  a  string  connecting 
the  joints  B  and  E.     If  the  weight  of  each  bar  be    W,  show  that  the 

W 

tension  of  the  string  is  —   (tan  54°  +  3  tan  18°),  and  find  the  magni- 
tudes and  directions  of  the  stresses  at  the  joints. 

5.  Six  bars  of  equal  length  and  weight  (  =  IV},  freely  articulated  at 
the  extremities,  form  a  regular  hexagon  ABCDEF. 

First,  if  the  system  hang  in  a  vertical  plane,  the  bar  AB  being  fixed 
in  a  horizontal  position,  and  the  form  of  the  hexagon  being  preserved 
by  means  of  a  string  connecting  the  middle  points  of  AB  and  DE,  show 

W 
that  (a)  the  tension  of  the  string  is  3  W,  (b)  the  stress  at  C  is  — —  and 

horizontal,  (c}  the  stress  at  D  is  W  4/il  and  makes  with  the  vertical 

12 

an  angle  cot  -1  2  1/3. 

Second,  if  the  system  rest  in  a  vertical  plane,  the  bar  DE  being  fixed 
in  a  horizontal  position,  and  the  form  of  the  hexagon  being  preserved 


EXAMPLES.  71 

by  means  of  a  string  connecting  the  joints    C  and   F,   show   that   (a) 

the  tension  of  the  string  is  W  — --,  (b)  the  stress  at  C  is  W  A/ 21.  and 

V3  12 

makes  with  C#  an  angle  sin  -'/f/JL,  (V)  the  stress  at  B  is    £F  4/-Z. 

124  12 

and  makes  with  CD  an  angle  sin  ~1-i/  — • 

Third,  if  the  system  hang  in  a  vertical  plane,  the  joint  ^  being  fixed, 
and  the  form  of  the  hexagon  being  preserved  by  means  of  strings  con- 
necting A  with  the  joints  E,  D,  and  C,  show  that  (a)  the  tension  of 
each  of  the  strings  AE  and  AC  is  W  V^ ,  (6)  the  tension  of  the  string 
AD  is  2  W;  and  determine  the  magnitudes  and  directions  of  the  stresses 
at  the  joints,  assuming  that  the  strings  are  connected  with  pins  distinct 
from  the  bars. 

6.  Show  that  the  stresses  at  C  and  F  in  the  first  case  of  Ex.  5  remain 
horizontal  when  the  bars  AF,  FE,  BC,  CD  are  replaced  by  any  others 
which  are  all  equally  inclined  to  the  horizon. 

7.  If  the  pole  of  a  funicular  polygon  describe  a  straight  line,  show 
that  the  corresponding  sides  of  successive  funicular  polygons  with  re- 
spect to  successive  positions  of  the  pole  will  intersect  in  a  straight  line 
which  is  parallel  to  the  locus  of  the  pole. 

8.  A  system  of  heavy  bars,  freely  articulated,  is  suspended  from  two 
fixed  points;  determine  the  magnitudes  and  directions  of  the  stresses  at 
the  joints.     If  the  bars  are  all  of  equal  weight  and  length,  show  that  the 
tangents  of  the  angles  which  successive  bars  make  with  the  horizontal 
are  in  arithmetic  progression. 

9.  If  an  .even  number  of  bars  of  equal  length  and  weight  rest  in  equi- 
librium in  the  form  of  an  arch,  and  if  a\,  a2,  .  .  .  an  be  the  respective 
angles  of  inclination  to  the  horizon  of  the   ist,  2d,  .  .  .  ;zth  bars  count- 
ing from  the  top,  show  that 

27Z   +  I 

tanar«-i.j  = tan  aH. 

2;z  —  I 

10.  Three  bars,  freely  articulated,  form  an  equilateral  triangle  ABC. 
The  system  rests  in  a  vertical  plane  upon  supports  at  B  and  C  in   the 
same  horizontal  line,  and  a  weight  W  is  suspended  from  A.     Determine 
the  stress  in  BC,  neglecting  the  weight  of  the  bars. 

W 

Ans.   -. 

2V3 

11.  Three  bars,  freely  articulated,  form  a  triangle  ABC,  and  the  sys- 
tem is  kept  in  equilibrium  by  three  forces  acting  on  the  joints.     Deter- 
mine the  stress  in  each  bar. 

What  relation  holds  between  the  stresses  when  the  lines  of  action  of 


72  THEORY  OF  STRUCTURES. 

the   forces   meet  (a)   in   the   centrpid,    (b)    in   the   orthocentre   of  the 
triangle  ? 

12.  A  triangular  truss  of  white  pine  consists  of  two  equal  rafters  AB, 
AC,  and  a  tie-beam  BC\  the  span  of  the  truss  is  30  ft.  and  its  rise  is  7$ 
ft.  ;  the  uniformly  distributed  load  upon  each  rafter  is  8400  Ibs.  Deter- 
mine the  stresses  in  the  several  members. 

Ans.  Stress  in  BC  '—  8400  Ibs.,  in  AB  =  4200  4/5  Ibs. 

33.  ABCD  is  a  quadrilateral  truss,  AB  and  CD  being  horizontal  and 
15  and  30  ft.  in  length,  respectively.  The  length  of  AC  is  10  ft.,  and  its 
inclination  to  the  vertical  is  60°.  A  weight  WL  is  placed  at  C,  and  Wi 
at  D.  What  must  be  the  relation  between  W\  and  Wi  so  that  the  truss 
may  not  be  deformed  ?  For  any  other  relation  between  W\  and  W*  , 
explain  how  you  would  modify  the  truss  to  prevent  deformation,  and 
find  the  stresses  in  all  the  members. 

Ans.    Wi  ssrW^J-Lti. 

2 

14.  A  Warren  girder  80  ft.  long  is  formed  oifive  equilateral  triangles. 
Weights  of  2,  3,  4,  5,  tons  are  concentrated,  respectively,  at  the  ist,  2d, 
3d,  and  4th  apex  along  the  upper  chord.     Determine  the  stresses  in  all 
the  members  of  the  girder. 

Ans.—  Tension  Chord  :  Stress  in  ist  bay  =  2  4/3  ;  2d  =  5$-  4/3  ; 

3d  =  7  Vl  ;  4th  =  6i  V~Z  ;  5th  =  2f  V~3> 
Compression  Chord  :  Stress  in  ist  bay  =  41/3  ; 

2d  =  6f  4/3  ;  3d  =  7-^3  ;  4th  =  5^-4/3. 
Diagonals  :  Stress  in  ist  and  2d.  =44/3; 

3d  and  4th  =  2$  4/3  ;  5th  and  6th  =  f  4/3  : 

7th  and  8th  =24/3;  9th  and  loth  =  5i  4/3- 

15.  In  a  quadrilateral  truss  ABCD,  AD  is  horizontal,  AB  and  BC  are 
inclined  at  angles  of  60°  and  30°  respectively  to  the  horizontal,  and  CD 
is  inclined  at  45°  to  the  horizontal.     What  weight  must  be  concentrated 
at  Cto  maintain  the  equilibrium  of  the  frame  under  a  weight  JV  at  £J 

If  a  weight  W  is  placed  at  Cas  well  as  at  D,  what  member  must  be 
introduced  to  prevent  distortion?  What  will  be  the  stress  in  that 
member  ? 

Ans.  First: 


Second:  Introduce  braced/?  and  let  BDA  =  a. 

Then  stress  in  ^=Jg2-/"3)    . 

2  sin  (60    +  (x) 


EXAMPLES.  73 

1  6.  The  boom  AB  of  the  accompanying  truss  is  supported  at  five 
intermediate   points  dividing  the   length 
into  six  segments  each  10  ft.  long.      The 


depth  of  the  truss  =  10  ft.      Draw  stress     ^  ™^\//^   h 
diagrams  for  the  following  cases  :  FlG-  I24- 

(a)  A  weight  of  100  Ibs.  at  each  intermediate  point  of  support. 
(&)   Weights  of  100,  200,  300,  400,  500  Ibs.       in  order  at  these  points. 
Ans.  (a)  Stress  in  a  =  375  ;  b  =  325  ;  c  =  375  ;  //  =  450  ; 
m  =  125  1/?3  ;  n  =  50  1/5  ;  o  -  504/5  ; 
p  =  25  4/73  Ibs. 

(£)  Stress  in  «  =  875  ;  b  —  825;  ^  =  925;  k  —1350;  d=  1325, 
*?  =  1125;  7  =  1375;  m  =  $oy$;  w  =  1004/5; 
0  =  141!  1/?3  ;  ^  =  8£|/?3  ;  r  =  200^5  ; 
s  =  250  1/5  ;  /  =  458^  4/?3  Ibs. 

17.  The  rafters  AB,  AC  of  a  factory  roof  are  18  and  24  ft.  in  length 
respectively.  The  tie  BC  is  horizontal  and  30  ft.  long.  The  middle 
points  of  the  rafters  are  supported  by  struts  DE,  DF  from  the  middle 
point  D  of  the  tie  BC\  the  point  D  is  supported  by  the  tie-rod  AD. 
The  truss  carries  a  load  of  500  Ibs.  at  each  of  the  points  E,  A,  and  F. 
Find  the  stresses  in  all  the  members.  Secondly,  find  the  stresses  in  the 
members  when  the  rafter  AB  is  subjected  to  a  normal  pressure  of  300 
Ibs.  per  lineal  ft.,  rollers  being  at  C. 

Ans.  Stress  in  BE  =  1  1  12^  ;  EA  =  800  ;  CF=  ioi6f  ;  FA  =  600  ; 
BD  =  667i  ;  CD  =  813*  ;  DE  =  312*  ;  DF  =  416!; 
AD  =  502  Ibs. 

Stresses  due  to  300  Ibs.  in  BE  —  1012^  ;  EA  —  1800  ; 
DE  =  2812^  ;  BD  =  3847^  ;  AD  =  2250  ; 
DC  =  2160  ;  v4C  =  2700  ;  Z>.F  =  o. 

1  8.  If  it  be  assumed  in  the  first  part  of  the  last,  question  that  the 
whole  of  the  weight  is  concentrated  at  the  points  E  and  F,  draw  the 
stress  diagram. 

19.  A  triangular  truss  consists  of  two  equal  rafters  AB,  AC  and  a  tie- 
beam  BC,  all  of  white  pine  ;  the  centre  D  of  the  tie-beam  is  supported 
from  A  by  a  wrought-iron  rod  AD;  the  uniformly  distributed  load  upon 
each  rafter  is  8400  Ibs.,  and  upon  the  tie-beam  is  36000  Ibs.  ;  determine 
(a)  the  stresses  in  the  different  members,  BC  being  40  ft.  and  AD  20  ft. 
What  (b)  will  be  the  effect  upon  the  several  members  if  the  centre  of  the 
tie-beam  be  supported  upon  a  wall,  and  if  for  the  rod  a  post  be  substi- 
tuted against  which  the  heads  of  the  rafters  can  rest?  Assume  that  the 
pressure  between  the  rafter  and  post  acts  at  right  angles  to  the  rafter. 
Ans.  (a)  Stresses  in  BD  =  13200;  AD  =  18000;  AB  =  13200  |/2  Ibs. 
(0)  "  "  =  4200;  "  =  8400;  "  =  6300  |/2~lbs. 


74  THEORY  OF  STRUCTURES. 

20.  A  triangular  truss  of  white  pine  consists  of  a  rafter  AC,  a  vertical 
post  AB,  and  a  horizontal  tie-beam  BC\  the  load  upon  the  rafter  is  300 
Ibs.  per  lineal  foot ;  AC  =  30  ft.,  AB  =  6  ft.     Find  the  resultant  pressure 
at  C. 

Ans.  4409  Ibs. 

Find  the  stresses  in  the  several  members  when  the  centre  D  of  the 
rafter  is  also  supported  by  a  strut  from  B. 

Ans.  Stress  in  BC  =  4500  |/6;  CD  =  22500;  DB  —  11250; 
DA  —  1 1 250 ;  AB  =  2250  Ibs. 

21.  The  rafters  AB,  AC  of  a.  roof-truss  are  20  ft.  long,  and  are  sup- 
ported at  the  centres  by  the  struts  DE,  DF;  the  centre  D  of  the  tie- 
beam  BC  is  supported  by  a  tie-rod  AD,  10  ft.  long;  the  uniformly  dis- 
tributed load  upon  AB  is  8000  Ibs.,  and  upon  AC  is  2400  Ibs.    Determine 
the  stresses  in  all  the  members. 

What  will  be  the  effect  upon  the  several  members  if  AB  be  subjected 
to  a  horizontal  pressure  of  156  Ibs.  per  lineal  foot? 

Ans.  (a)  Stress  in  BD  =  4600  1/3  ;  BE  =  9200 ;  EA  =  5200 ; 
ED  =  4000 ;  AD  —  2600  ;  DF  =  1 200 ; 
AF  =  5200  ;    CF  =  6400  ;    CD  =  3200  1/3. 
(£)  Tens,  in  BE  =  520  4/3 ;  AD  =  260  1/3  ;  compres.  in 

ED  —  520  1/3";   AC  =  520  j/J;  DC=  780. 
No  stresses  in  BD,  AE. 

22.  Determine  the  stresses  in  all  the  members  of  the  truss  in  the 
preceding  question,  assuming  the  tie-beam  to  be  also  loaded  with  a 
weight  of  600  Ibs.  per  lineal  foot. 

Ans.  Stress  in  AB  increased  by  6000  4/3  Ibs. ;  in  BC  by  9000  Ibs.; 
in  AD  by  6000  4/3  Ibs. 

23.  A  horizontal  beam  is  trussed  and  supported  by  a  vertical  strut  at 
its  middle  point.     If  a  loaded  wheel  roll  across  the  beam,  show  that  the 
stress  in  each  member  increases  proportionately  with  the  distance  of  the 
wheel  from  the  end. 

W  Wx 

Ans.  Stress  in  tie-beam  (hor.)  =  —rx  cot  0;    on  tie  =  -,— —  ; 

/  /sin  0 

W 
%  on  strut  =  —fix. 

24.  A  frame  is  composed  of  a  horizontal  top-beam  40  ft.  long,  two 
vertical  struts  3  ft.  long,  and  three  tie-rods  of  which  the  middle  one  is 
horizontal  and  15  ft.  long.     Find  the  stresses  produced  in  the  several 
members  when  a  single  load  of  6000  Ibs.  is  concentrated  at  the  head  of 
each  strut. 

Ans.  Stress  in  horizontal  members  =  50000  Ibs. 
"        "   sloping  "          =  51420    " 

"       "   struts  =  12000    " 


EXAMPLES,  75 

25.  If  a  wheel  loaded  with  12000  Ibs.  travel  over  the  top-beam  in  the 
last  question,  what  members  must  be  introduced  to  prevent  distortion  ? 
What  are  the,  maximum  stresses  to  which  these  members  will  be  sub- 
jected ? 

Ans.  19122  Ibs. 

26.  A  beam  of  30  ft.  span  is  supported  by  an  inverted  queen-truss, 
the  queens  being  each  3  ft.  long  and  the  bottom  horizontal  member  10 
ft.  long.     Find  the  stresses  in  the  several  members  due  to  a  weight  W 
at  the  head  of  a  queen,  introducing  the  diagonal  required  to  prevent 
distortion.     Also  find  the  stresses  due  to  a  weight  W  at  centre  of  beam. 

20  ^o 

Ans.—  i.  Stress  \nAB  =  —  W\  AE=2.^2lV\  EF=~W; 

BE=-W\     BF  =  i.i6W\  BC  =  —W\ 

=  \.\6W. 

W 


2.  Stress  \*AB  = 

-W. 

27.  A   roof-truss  of   20  ft.  span  and  8  ft.  rise  is  composed  of   two 
rafters  and  a  horizontal  tie-rod  between  the  feet.     The  load  upon  the 
truss  =  500  Ibs.  per  foot  of  span.     Find  the  pull  on  the  tie.     What  would 
the  pull  be  if  the  rod  were  raised  4  ft.? 

Ans.  3125  Ibs.  ;  6250  Ibs. 

28,  The  rafters  AB,  AC  of  a  roof  are  unequal  in  length  and  are  in- 
clined at  angles  a,  ft  to  the  vertical  ;  the  uniformly  distributed  load  upon 
AB  =  W\  ,  upon  AC  '  =  Wi  .     Find  the  tension  on  the  tie-beam. 

+  Wi     sin  a    sin  ft 


Ans. 


sin  (a  +  ft}' 


29.  In  the  last  question,  if  the  span  =  10  ft.,  a  =  60°  and  ft  —  45°,  find 
the  tension  on  the  tie,  the  rafters  being  spaced  2^  ft.  centre  to  centre, 
and  the  roof-load  being  20  Ibs.  per  square  foot. 

Ans.  198  Ibs. 

30.  The  equal  rafters  AB,  AC  for  a  roof  of  10  ft.  span  and  2\  ft.  rise 
are  spaced  2\  ft.  centre  to  centre ;  the  weight  of  the  roof-covering,  etc. 
=  20  Ibs.  per  square  foot.    Find  the  vertical  pressure  and  outward  thrust 
at  the  foot  of  a  rafter. 

Ans.  Total  vertical  pressure  =  125  4/5  Ibs.  =  horizontal  thrust. 

31.  The  lengths  of  the  tie-beam  and  two  rafters  of  a  roof-truss  are  in 
the  ratios  of  5  :  4  :  3.     Find  the  stresses  in  the  several  members  when 
the  load  upon  each  rafter  is  uniformly  distributed  and  equal  to  100  Ibs. 

Ans.  Stress  in  tie  =  48  Ibs.  ;  in  one  rafter  =  60  Ibs.;  in  other  =  80  Ibs. 


/  THEORY  OF   STRUCTURES. 

32.  In  a  triangular  truss  the  rafters  each  slope  at  30° ;  the  load  upon 
the  apex  =  100  Ibs.     Find  the  thrust  of  the  roof  and  the  stress  in  each 
rafter. 

Ans.  100  Ibs.;  86.6  Ibs. 

33.  A  roof-truss  is  composed  of  two  equal  rafters  and  a  tie-beam,  and 
the  span  =  4  times  the  rise;  the  load  at  the  apex  =  4000  Ibs.     Find  the 
stresses  in  the  several  members. 

Secondly,  if  a  man  of  150  Ibs.  stands  at  the  middle  of  a  rafter,  by 
how  much  will  the  stress  in  the  tie-beam  be  increased  ? 

Ans. — i.  Stress  in  tie  =  4000  Ibs. ;    in  each  rafter  =  2000  1/5  Ibs. 
2.  75  Ibs. 

34.  A  king-post  truss  for  a  roof  of  30  ft.  span  and  7^  ft.  rise  is  com- 
posed of  two  equal  rafters  AB,  AC,  the  horizontal  tie-beam   BC,  the 
vertical  tie  AD,  and  the  struts  DE,  DF  from  the  middle  point  D  of  the 
tie-beam  to  the  middle  points  of  the  rafters ;  the  roof-load  =  20  Ibs.  per 
square  foot  of  roof-surface,  and  the  rafters  are  spaced   10  ft.  centre  to 
centre.     Find  the  stresses  in  the  several  members. 

Second,  find  the  altered  stresses  when  a  man  of  150  Ibs.  weight  stands 
on  the  ridge. 

Third,  find  the  altered  stresses  when  the  tie-beam  supports  a  celling 
weighing  12  Ibs.  per  square  foot. 

Ans. — i.  Stress  in  BE  =  56250  Ibs. ;  BD  =  2250  4/5  Ibs.; 
AE  —  46875  Ibs. ;  DE  =  9375  Ibs.; 
AD=  7500  4/5"  Ibs. 

2.  Stresses  in  BD,  BE,  AE  increased  by  150  Ibs.,  75  1/5 

Ibs.,  and  75  \' '5  Ibs.,  respectively  ;  other  stresses  un- 
changed. 

3.  Stresses  in  AD,  tie-beam,  and  rafters  increased  by  1800, 

1800.  and  900  1/5  Ibs.,  respectively  ;  other  stresses 
unchanged. 

35.  The  platform  of  a  bridge  for  a  clear  span  of  60  ft.  is  carried  by 
two  queen-trusses  15  ft.  deep;  the  upper  horizontal  member  of  the  truss 
is  20  ft.  long;  the  load  upon  the  bridge  =  50  Ibs.  per  square  foot  of  plat- 
form, which  is  12  ft.  wide.     Find  the  stresses  in  the  several  members. 

Ans.  Stress    in  vertical  ^=  6000   Ibs.;    in  each  sloping  member 
=  loooo  Ibs. ;  in  each  horizontal  member  =  8000  Ibs. 

36.  If  a  single  load  of  6000  Ibs.  pass  over  the  bridge  in  the  last  ques- 
tion, and  if  its  effect  is  equally  divided  between  the  trusses,  find  (a)  the 
greatest  stress  in  the  members  of  the  truss,  and  also  (lij  in  the  members 
which    must   be   introduced  to  prevent  distortion.      Also  find   (c)    the 
stresses  when  one  half  the  bridge  carries  an  additional  load  of  50  Ibs.  per 
square  foot  of  platform. 


EXAMPLES.  77 

Ans.— (a)   In  sloping  end  strut  =  3333*  Ibs.;  horizontal  tie  = 

2666f  Ibs.;  horizontal  strut  =  1333*  Ibs. 
(<:)  In  sloping  end  strut  =  6250  Ibs.;   horizontal  tie  = 

5000  Ibs.;  horizontal  strut  =  3000  Ibs. 
(b)  In  case  (a)  =  i666f  Ibs.;  in  case  (c)  =  2500  Ibs. 

37.  A  roof-truss  consists  of  two  equal  rafters  AB,  AC  inclined  at  60° 
to  the  vertical,  of  a  horizontal  tie-beam  BC  of  length  /,  of  a  collar-beam 

DE  of  length  — ,  and  of    queen-posts    DF,  EG  at   each  end  of    the 

collar-beam  ;  the  truss  is  loaded  with  a  weight  of  2600  Ibs.  at  the  vertex, 
a  weight  of  4000  Ibs.  at  one  collar-beam  joint,  a  weight  of  1200  Ibs.  at 
the  other,  and  a  weight  of  1500  Ibs.  at  the  foot  of  each  queen;  the 
diagonal  DG  is  inserted  to  provide  for  the  unequal  distribution  of  load. 
Find  the  stresses  in  all  members. 

Ans.     Stress  in  BD=  11733*;  BF  =  5866f  4/3  ;  DF  =  1 500  ; 
DA  =  2600  ;  DE  =  3633*  4/3  ;  DG  =  i866f ; 
GC  =  4933i  1/3  ;  GE  =  2433* ;  CE  =  9866f  ; 
AE  —  2600  Ibs. 

38.  The  rafters  AB,  AC  are  supported  at  the  centres  by  the  struts 
DE,  DF;   the  centre    of  the  tie-beam  is   supported    by   the   tie   AD\ 
BC  =  30  ft.,  AD  =  7i  ft.  ;  the  load  upon  AB  is  4000  Ibs.,  that  upon  AC 
1600  Ibs.     Find  the  stresses  in  all  the  members.     By  an  accident  the 
strut  DE  was  torn  away;  how  were  the  stresses  in  the  other  members 
affected  ? 

Ans.— Case  i :  Stress  in  BE  —  2400  4/5";  BD  =  4800  ; 

DE  =  icoo  4/5";  AE  =  1400  4/5  ; 
AF  =  1400  4/5";  DF  =  400  V5  ; 
FC  =  1800  i/J";  DC  —  3600  Ibs. 
Case  2:  Stress  in  BA  =  1400  4/5";  BD  =  2800  ; 
AD  =  400 ;  AF  =  1400  4/5  ; 
FC  =  1800  4/5";  DF  =  400  4/5"; 
DC  =  3600. 

39.  The  platform  of  a  bridge  for  a  clear  span  of  60  ft.  is  carried  by 
two  trusses  15  ft.  deep,  of  the  type  shown  by  the  p       n 
accompanying    diagram  ;     the     load     upon    the 

bridge  is  50  Ibs.  per  square  foot  of  platform, 
which  is  12  ft.  wide.  Find  the  stresses  in  the 
several  members. 

Ans.     Stress  in  BE  =  13500;  ^£  =  67504/5";  EG  =  4000 

ED  =  1 3500 ;  CD  =  2250  \/~^\  GA  =  4500 

AD  =  9000  Ibs. 


78  THEORY  OF  STRUCTURES. 

40.  If  a  single  weight  of  2000  Ibs.  pass  over  a  truss  similar  to  that 
shown  in  the  preceding  question,  find  the  stresses  in  the  several  members 
when  the  load  is  (i)  at  E,  (2)  at  D. 

Ans.—Case  i:  Stress  in  BG  =  1500  1/5  ;  BE  =  3000  ; 
EG  =  2000  ;  ED  —  3000  ; 
GD  =  looo  4/5  ;  AG  =  500  1/5  ; 

AH—  500^/5";  DH=o;  FH-o\ 
DF  —  1000  ;  FC  =  1000  ; 
£77=5004/5"  Ibs. 
Case  2  :  Stress  in  BA  and  CA  =  1000  1/5  ; 

BD  and  DC  =  2000  ;  AD  =  2000  Ibs., 
and  in  other  members  =  o. 

41.  A  white-pine  triangular  truss  consists  of  two  rafters  AB,  -AC,  of 
unequal  length,  and  a  tie-beam  BC.     A  vertical  wrought-iron  rod  from 
A,  10  ft.  long,  supports  the  tie-beam  at  a  point  D,  dividing  its  length 
into  the  segments  BD  =  ioit.  and  CD  =  20  ft.     The  load  upon  each 
rafter  is  300  Ibs.  per  lineal  ft.  ;  the  load  upon  the  tie-beam  is  18,000  Ibs., 
uniformly  distributed.     Determine  the  stresses  in  the  several  members. 
Ans.     In  AB  =  9650  V  2"  Ibs.  ;  AC=  4825  4/5  Ibs.  ;  BD  =  CD  =  9650  Ibs. 

42.  The  post  of  a  jib-crane  is  10  ft.  ;  the  weight  lifted  =  W\  the  jib 
is  inclined  at  30°,  and  the  tie  at  60°,  to  the  vertical.     Find  (a)  the  stresses 
in  the  jib  and  tie,  and  also  the  B.  M.  at  the  foot  of  the  post. 

How  (b}  will  these  stresses  be  modified  if  the  chain  has  four  falls,  and 
if  it  passes  to  the  chain-barrel  in  a  direction  bisecting  the  angle  between 
the  jib  and  tie  ? 

Ans.—  (a)  Stress  in  tie  =  W\  in  jib  =  JFy  J.  B.  M.  =  5  4/3"  ft.  tons. 


43.  An  ordinary  jib-crane  is  required  to  lift  a  weight  of  10  tons  at  a 
horizontal  distance  of  9  ft.  from  the  axis  of  the  post.  The  hanging  part 
of  the  chain  is  in  four  falls;  the  jib  is  15  ft.  long,  and  the  top  of  the 
post  is  16^  ft.  above  ground.  Find  the  stresses  in  the  jib  and  tie  when 
the  chain  passes  (i)  along  the  jib,  (2)  along  the  tie. 

The  post  turns  round  a  vertical  axis.  Find  the  direction  and  magni- 
tude of  the  pressure  at  the  toe,  which  is  3  ft.  below  ground. 

Ans.  —  (i)  Stress  in  tie  =  3°_1_5  tons;  in  jib  =  ii£f  tons. 


ii 


(2) 


=  [3°  ^5  __  2^)  tons  ;  in  jib  =  9TaT  tons. 
.    \     ii  / 


Pressure  on  toe  =  10  \/io  tons,  and  is  inclined  to  vertical  at  an  angle 

tan  ~  l  i. 


EXAMPLES.  79 

44.  In  the  crane  represented  by  the  figure  AB  =  AC  =  A 
35ft.;  2?C=2oft.;  £D=2oi\..\  the  weight  lifted  =  25  tons; 
y4C  slopes  at  45°  ;    the  chain  hangs  in  four  falls  and  passes 
from  A  to  D.    Find  the  stresses  in  all  the  members  and 
the  upward  pull  at  D. 

Ans.     Stress  in  £C=26;  AC  =  47.6  ;  AB  =  28.4  ;  CD  =  32.8  tons. 
Vertical  pull  at  D  =  31.3  tons. 

45.  The  figure  represents  the  framing  of  an  hydraulic  crane.     AB.—BD 

=  DF  =  FG=^  HK  =  5  ft.;  KG  =  BC  =  2*-  ft.    Find  the 

Hf\~7^\—  ^L      stresses  in  the  members  of  the  crane  when  the  weight 

KG  FOB"  A  (i  ton)  lifted  is  (a)  at  A\(&)a.\.B\  (c)  at  D.     Also  (d)  find 

FIG.  127.        the  stresses  when  there  is  an  additional  weight  of  £  ton 

at  each  of  the  points  B,  D,  F,  and  G. 

Ans—  (a)  Stress  in  tons  in  AB  =  BD  —  2  ;  DF  =  FG  =       ; 


9009 
Iy 


Stress  in  tons  in  ^^  =  o  =  ^C;  ^C=  i  ;  CE  =       |/2  ; 


=FG  =       ;  GK  = 

13  II  2 


M3 
Stress  in  AB  =  o  = 


*-£«<» 


=  7^317; 


80  THEORY  OF  STRUCTURES. 


(d)  Stress  in  tons  in  AB  =  o  =  AC;  EC  =  -  =  EF ; 


46.  The  inclined  bars  of  the  trape- 
zoidal truss   represented  by  the  figure 
make  angles  of  45°  with  the  vertical ;  a 
load  of  10  tons  is  applied  at  the  top 
joint  of  the  left  rafter  in  a  direction  of 
^     45°  with  the  vertical.      Assuming  the 
W  "P         reaction  at  the  right  to  be  vertical,  find 

FlG>  I28>  the  stresses  in  all  the  pieces  of  the  frame. 

Ans.    Vert,  reaction  at  D—   -  yT;   stress  in  DE=  —  |/2~; 

•j  3 

,.  0  20     /-  10     ,— 

3  3 

AC  =  I0  ;  CE  =  -  1/2"  tonr. 

47.  The  post  of  a  derrick-crane  is  30  ft.  high  ;  the  horizontal  traces  of 
the  two  back-stays  are  at  right  angles  to  each  other,  and  are  15  ft.  and 
25  ft.  in  length.    Show  that  the  angle  between  the  shorter  trace  and  the 
plane  of  the  jib  and  tie,  when  the  stress  in  the  post  is  a  maximum,  is 

30°  58'. 

Also  find  the  greatest  stresses  in  the  different  members  of  the  crane 
when  the  jib,  which  is  50  ft.  long  and  is  hinged  at  the  foot  of  the  post,  is 
inclined  at  45°  to  the  vertical,  the  weight  lifted  being  4000  Ibs. 

Ans.  Stress  in  jib  =  6666f  Ibs.;  in  tie  =4768. 4  Ibs.;  max. 
thrust  along  post  =  10991.5  Ibs.  ;  max.  stress  on  long 
back-stay  =  7362.7  Ibs.;  on  short  back-stay  =  10539  Ibs. 

48.  A  queen-truss  for  a  roof  consists  of  two  horizontal  members,  the 
lower  48  ft.  long,  the  upper  16  ft.  long;  two  inclined  members  AB,  DC, 
and  two  queens  BE,  CF,  each  8  ft.  long;  the  points  E,  ^divide  AD  into 
three  equal  segments ;   the   load   upon   the   members  AB,  BC,  CD  is 
120  Ibs.  per  lineal  foot.     Find  (a)  the  stresses  in  the  several  members. 
How  (b)  will  these  stresses  be  modified  if  struts  are  introduced  from  the 


EXAMPLES.  Si 

feet  of  the  queens  to  the  middle  points  G,  H  of  the  inclined  members  ? 
In  this  latter  case  also,  determine  (c)  the  stresses  due  to  a  wind-pressure 
of  120  Ibs.  per  lineal  ft.  normal  to  AB,  assuming  that  the  horizontal  re- 
action is  equally  divided  between  the  two  supports  at  A  and  D. 

Ans.—(a)  Stress  in  Ibs.  in  AE  —  4066.56  =  EF-DF  =  BC\ 

AB  =  4546.  $6  =  CD;  BE  =  2033.  28  =  CF. 
(£)  Stress  in  Ibs.  in  AE—  5139.84  =  DF\ 

BC  =  4066.56  =  AF;  AG  =  5746.56  -  DH\ 
BG  =  4546.56  =  CH\  EG  =  1200  =  FH\ 
BE^  536.64  =  CF. 

(c)  Additional  stress  in  AG  =  1040.4/5";  BG  —  680  4/5"; 
GE-  600  4/5";  ^£=2320  ;  BE  =  600  ;  BC=  400  4/5"; 
BE=  400  4/5";  CF  =  400  ;  CB  =  400  4/5";  EF  =  1  120  ; 


(In  case  (<:)  the  brace  BF  is  introduced  to  prevent  distortion.) 

49.  A  pair  of  shear-legs,  each  25  ft.  long,  with  the  point  of  suspension 
20  ft.  vertically  above  the  ground  surface,  is  supported  by  a  tie  100  ft. 
long  ;  distance  between  feet  of  legs  =  10  4/5  ft.     Find  the  thrusts  along 
the  legs  and  the  tension  in  the  tie  when  a  weight  of  2  tons  is  being 
lifted. 

Ans.  Tension  in  tie  =  1.137  tons;  compn.  in  each  leg  =  1.87  tons. 

50.  In  the  crane  ABC,  the  vertical  post  AB  =  15',  the  jib  AC  '=  23', 
and  the  angle  BAC  '=  30°.     Find  (a)  the  stresses  in  the  jib  and  tie,  and 
also  the  bending  moment  at  the  foot  of  the  post  when  the  crane  lifts  a 
weight,  of  4  tons. 

The  throw  is  increased  by  adding  two  horizontal  members  CE,  BD 
and  an  inclined  member  DE,  the  figure  BE  being  a  parallelogram  and 
the  diagonal  CD  coincident  in  direction  with  CA.  Find  (b)  the  stresses 
in  the  several  members  of  the  crane  as  thus  modified,  the  weight  lifted 
being  the  same. 

In  the  latter  case  show  (c)  how  the  stresses  in  the  members  are  affected 
when  the  chain,  which  is  in  four  falls,  passes  from  E  to  B  and  then  down 
the  post. 

Ans.  —  (a)  Tension  in  tie  =  3^  tons;    thrust  in  jib  =  6T27  tons; 

(b)  Stress  in  CE  =  9.34  ;  in  ED  =  10.16;  in  CB  =  13.49; 

in  CD  =  6.15;  in  DA  =  10.7  ;    in  BD  =  7  tons. 

(c)  Stress  in  CE  =  8.9;    in  ED  =  10.7;    in    CB  =  12.9; 

in  CD  =  5.8  ;  in  DA  —  10.7  ;  in  BD  =  7.4  tons. 

51.  The  horizontal  traces  of  the  two  back-stays  of  a  derrick-crane 
are  x  and  y  feet  in  length,  and  the  angle  between  them  is  fi.     Show  that 

cos  (ft  —  6)      x 

the  stress  in  the  post  is  a  maximum  when  -  -  —  =—0  beine  the 

cos  6  y 

angle  between  the  trace  x  and  the  plane  of  the  jib  and  tie. 


82 


THEORY  OF  STRUCTURES. 


52.  The  two  back-stays  of  a  derrick-crane  are  each  38'  long,  and  the 
angle  between  their  horizontal  traces  2  tan-IT5^;  height  of  the  crane- 
post  =  32';  the  length  of  the  jib  =  40' ;  the  throw  of  the  crane  =  20' ; 
the  weight  lifted  =  4  tons.     Determine  the  stresses  in  the  several  mem- 
.bers  and  the  upward  pull  at  the  foot  of  each  back-stay  when  the  plane  of 
the  jib  and  post  (a)  bisects  the  angle  between  the  horizontal  traces  of 
xhe  back-stays,  (b)  passes  through  a  back-stay. 

Ans.   In  jib  =  5  ;  in  tie  =  2.52  tons  ;  in  back-stay  in  (a)  =  2.56, 
in  (b)  —  4.7  tons. 

53.  Find   the   stresses   in   the   members   of  the   crane   represented 
E          35'         B  by    the    figure;    also    find    balance-weight 

atC 

Ans.  Stress  in  BE  =  25  ;  DE  =  26.9; 
DB  =  21.08;  DA  =  26.08; 
BA  =  .24;  BC  =  1 8. 12  tons. 
Counterweight  at  C=  15.14 
tons. 

54.  Draw  the  stress  diagram  for  the  truss  represented  by  the  figure, 
the  load  at  each  of  the  points  £  and  C  being  500  Ibs. 


FIG.  130. 

Also,  if  the  rafter  AB  is  subjected  to  a  nominal  wind-pressure  of  100 
Ibs.  per  lineal  ft.,  introduce  the  additional  member  required  to  prevent 
deformation,  and  state  in  Ibs.  the  stress  it  should  be  designed  to  bear. 
Draw  the  stress  diagram  of  the  modified  truss,  assuming  that  the  foot  A 
is  fixed,  and  that  there  are  rollers  at  D. 
(AB  =  AE  =15';  BC  =  10' ;  angle  BAD  =  45° ;  angle  EAD  —  30°.) 

55.  The  post  AB  of  a  jib-crane  is 
20  ft. ;  the  jib  AC  is  inclined  at  30°  and 
the  tie  BC  at  45°  to  the  vertical ;  the 
weight  lifted   is    5    tons.      Find    the 
stresses   in   the  jib  and  tie  when  the 
chain  passes  (a)  along  the  jib,  (b}  along 
the  tie,  (c)  horizontally  from  C  to  the 
post. 

The  chain  has  two  falls. 

56.  In  a  mansard  roof  of  12  ft.  rise,  the  upper  triangular  portion  (of 
4  ft.  rise)  has  its  rafters  inclined  at  60°  to  the  vertical.     The  rafters  of  the 


FIG.  131. 


EXAMPLES.  83 

lower  portion  are  inclined  at  30°  to  the  vertical.  If  there  is  a  load  of 
1000  Ibs.  at  the  ridge,  find  the  load  at  each  intermediate  joint  necessary 
for  equilibrium,  and  the  thrust  of  the  roof. 

A  load  of  2000  Ibs.  is  concentrated  at  each  of  the  intermediate  joints 
and  a  brace  is  inserted  between  these  points.  Find  the  stress  in  the 
orace. 

Ans.   1000  Ibs. ;  thrust  =  500  4/3"  Ibs.;  333!-  |/J  Ibs. 

57.  The  horizontal  boom  CD  is  divided  into  eight  segments,  each 
8  ft.  long,  by  seven  intermediate  supports  ; 

the  depth  of  the  truss  at  each  end  =  16  ft.; 
a  weight  of  i  ton  is  concentrated  at  C  and 
at  D,  and  a  weight  of  2  tons  at  each  of  the 
points  of  division.  Determine  the  stresses 
in  the  several  members. 

58.  The  figure  is  a  skeleton  diagram  of  a  roof-truss  of  72  ft.  span  and 
12  ft.  deep ;  G,  K,  L,  O,  H  are  respectively  the  middle  points  of  AE,  EL, 

EF,  LF,  FB  ;  AE  =  EL  =  LF=  FB  =  20  ft.; 
the  trusses  are  12  ft.  centre  to  centre  ;  the  dead 


..  weight  of  the  roof  =  12  Ibs.  per  sq.  ft.;  the 

^s     normal  wind-pressure  upon  AE  may  be  taken 


FIG.  133.  _  30  jbg>  per  Sq    ft  .  the  encj  ^  js  fjxe(j  an(j 

B  is  on  rollers.     Draw  a  stress  diagram.     Show  by  dotted  lines  how  the 
stress  diagram  is  modified  with  rollers  under  A,  B  being  fixed. 

59.  The  platform  of  a  bridge  of  84  ft.  span  B  c  D  E 

and  9  ft.  deep  is  carried  by  a  pair  of  trusses  of       A^^Tn    P^P55^ 
the  type  shown  in  the  figure.     If  the  load  borne    ^ 
by  each  truss  is  300  Ibs.  per  lineal  ft.,  find  the  FIG.  134. 

stresses  in  all  the  members. 

Ans.     Stress  in  AB  =  6000  ;  AC  =  1200  4/73  ;  AD  =  3600  1/17  ; 

BC  —  4800  ;  CD  =  14400  ;  DE  =  28800. 
Stress  in  horiz.  chord  =  288000  ;  in  each  vertical  =  3600 
Ibs. 

60.  The  figure  represents  the  shore  portion  of  one  of  the  trusses  for 

a  cantilever  highway  bridge.     The  depth  of 
t   t  B 

truss  over  pier  =  51  ft.;  the  length  of  each 

panel  =  17  ft.;  the  load  at  A  (from  weight  of 


is.sooibs.     c><^j/p  centre  span)  —  16800  Ibs.;  the  width  of  road- 

way =  15  ft.;  the  load  per  sq.  ft.  of  roadway 
FIG.  135.  _  go  Ibs.     Find  the  stresses  in  all  the  mem- 

bers, assuming  the  reaction  at  the  pier  F  to  be  vertical. 

Ans.  ti  —ti  —  28000  ;  /3  =  36500  ;  /4  =  45000 ;  /5  =  53500  ; 

/6  =  55200  ;  ti  =  48400  ;  /8  =  41600  =  /»  ;    Ci  =  5600  1/34"; 
<:2  =  7300  V  34  ;  vi  =  10200  ;  vt  =  15300;  z/8  =  20400  ; 


84 


THEORY  OF  STRUCTURES. 


V*  =  25500 ;  7/5  =  45900  J  s/e  =  20400  ;  V-,  —  1 5300 ; 

1/6  =  10200  ;    <T3  =  9000  4/34  ;    <r4  =  10700  1/34  ; 

<r5  =  12400  4/34  ;    <r6  =  77500 ;    c-i  —  69000 ;    <r8  =  60500  ; 

<r9  =  52000  ;  di  =  1700  4/34  ;  </a  =  1700  4/6?  ; 

rt?3  =  1700  4/106  ;  </4  =  22100;  d*  =  1700  4/97"; 

dt  =  3400  4/13  ;  </7  =  8500  Ibs. 

61.  The  inner  flange  of  a  bent  crane  forms  a  quadrant  of  a  circle  of 
20  ft.  radius,  and  is  divided   into  four  equal  bays.     The  outer  flange 
forms  the  segment  of  a  circle  of  23  ft.  radius.     The  two  flanges  are  5  ft. 
apart  at  the  foot,  and  are  struck  from  centres  in  the  same  horizontal 
line.     The  bracing  consists  of  a  series  of  isosceles  triangles,  of  which  the 
bases  are  the  equal  bays  of  the  inner  flange.     The  crane  is  required  to 
lift  a  weight  of  10  tons.     Determine  the  stresses  in  all  the  members. 

62.  A  braced  semi-arch  is  10  ft.  deep  at  the  wall  and  projects  40  ft. 
The  upper  flange  is  horizontal,  is  divided  into  four  equal  bays,  and  carries 
a  uniformly  distributed  loatl  of  40  tons.     The  lower  flange  forms  the 
segment  of  a  circle  of  104  ft.  radius.     The  bracing  consists  of  a  series  of 
isosceles  triangles  of  which  the  bases  are  the  equal  bays  of  the  upper 
flange.     Determine  the  stresses  in  all  the  members. 

63.  The  domed  roof  of  a  gas-holder  for  a  clear  span  of  80  ft.  is  strength- 
ened by  secondary  and  primary  trussing  as  in  the  figure.     The  points  B 

and  C  are  connected  by  the  tie  BPC  passing 
beneath  the  central  strut  APt  which  is  15  ft. 
long,  and  is  also  common  to  all  the  primary 
trusses ;  the  rise  of  A  above  the  horizontal  is  5 
ft.;  the  secondary  truss  ABEF  consists  of  the 
equal  bays  AH,  HG,  GB,  the  ties  BE,  EF,  FA,  of  which  BE  is  horizon- 
tal, and  the  struts  GE,  FH,  which  are  each  2  ft.  6  in.  long  and  are  par- 
allel to  the  radius  to  the  centre  of  GH;  the  secondary  truss  ACLK  is 
similar  to  ABEF\  when  the  holder  is  empty  the  weight  supported  by 
the  truss  is  36000  Ibs.,  which  may  be  assumed  to  be  concentrated  at  G, 
H,  A,  M,  N,  in  the  proportions  8000,  4000,  1000,  4000,  and  8000  Ibs.,  re- 
spectively. Determine  the  stresses  in  the  different  members  of  the  truss. 

64.  The  figure  is  the  skeleton  diagram  of  a  cantilever  for  a  viaduct  in 

B 


91  tons       42  tons  42  tons  42  tone  21  tool 

FIG.  137. 


EXAMPLES. 


India.     Determine  graphically  the  stresses  in  the  various  members  un- 
der the  loading  indicated. 

65.  In  the  accompanying  roof-truss  AB  =  AC—  30  ft.,  and  the  struts 
ire  all  normal  to  the  rafters.     Find  the 

stresses  in  all  the  members,  the  load  at 
each  of  the  joints  in  the  rafters  being  2 

tons  (angle  ABC  =  30°  and  angle  DBC       B  ^y^LJ^J/^A^C 
=  10°).     How  will  the  stresses  be  mod- 
ified if  there  is  a  force  of  2  tons  acting 
at  each  of  the  points  of  support  between 

A  and  B  at  right  angles  to  the  rafter,  and  a  force  of  i  ton  at  A,  assum- 
ing that  the  end  B  is  fixed  and  that  C  rests  upon  rollers? 

66.  The  figure  represents  a  portion  of  a  Warren  girder  cut  off  by  the 

plane  MN  and  supported  upon  the  abutment  at 

A.    The  reaction  at  A  =20  tons  ;  the  load  con- 

/\     /\     /\    U   c      centrated  at  each  of  the  points  .#  =  4  tons.    Find 

A/    \/    V      vf  the  stresses  in  each  of  the  members  met  by  MN. 

B       B      Bj 

N  Ans.  Stress  in  tension  chord  =  —  ^  tons ; 

FIG.  139.  3 

in  compression  chord  =  32  1/3  tons ;  compression  in  diagonal  =  —  1/3 

tons. 

67.  The  figure  represents  a  portion  of  a  roof-truss  cut  off  by  a  plane 
MN  and  supported  at  A.     The  strut  DC  is 

vertical ;  AD  =  23  ft.,  and  the  distance  of  D 
from  AC=  7i  ft.;  the  angle  between  AC  and 
the  horizontal  =  cos-'£;  the  vertical  reac- 
tion at  A  —  7  tons ;  the  horizontal  reaction 
at  A  =  2|  tons ;  at  each  of  the  points  B  and 
C  a  weight  of  4  tons  is  concentrated.  Find 
the  stresses  in  the  members  met  by  MN. 
(AD  and  7\  make  equal  angles  with  the  FIG.  140. 

rafter.) 

Ans.  Ci  =  13.2  tons  ;   T*  =  2.1  tons  ;   7\  =  10.8  tons. 

68.  The  feet  of  the  equal  roof-rafters  AB,  A  Care  tied  by  rods  BDt 
CD  which  meet  under  the  vertex  and  are  joined  to  it  by  a  rod  AD.     If 
IVi ,  Wi  are  the  uniformly  distributed  loads  in  pounds  upon  AB,  AC, 

respectively,  and  if  5  is  the  span  of  the  roof  in  feet,  find  the  weight  of 
metal  (wrought-iron)  in  the  ties. 

Ans   -    — 5_ ?  5  cot  fi,    f  being  inch-stress  in  pounds,  and 

6          / 

ft  the  angle  ABD. 

(a)  If  AB  =  AC  =  20  ft.,  AD  —  5  ft.,  the   angle  BAD  =  60°,  find 


86  THEORY  OF  STRUCTURES. 

the  stresses  in  the  several  members  when  a  weight  of  3500  Ibs.  is  con- 
centrated at  the  vertex. 

Ans.  7000  Ibs.;  6309.8  Ibs.;  3500  Ibs. 

(b)  The  roof  in  (a)  is  loaded  with  10  Ibs.  per  square  foot  on  one  side 
and  33  Ibs.  per  square  foot  on  the  other;  the  trusses  being  13  ft.  centre 
to  centre.  Determine  (a)  the  stresses  in  the  several  members.  Examine 
(jj]  the  effect  of  a  horizontal  pressure  of  14  Ibs.  per  square  foot  on  the 
most  heavily  loaded  side,  assuming  that  the  reaction  is  equally  divided 
between  the  two  supports. 

Ans.   (a)  mSolbs.;  10077.65  Ibs.;  5590  Ibs. 

69.  In   the  truss  represented  in  the  accompanying  figure,  the  load 

on  AB  =  IVi ,  on  AC  =  W* ;  the  angle  ABD  —  ft ; 
AD  =  BD  =  AE  =  CE.  Find  the  total  weight 
of  metal  (wrought-iron)  in  the  tie-rods. 

5     Wi  +  Wi 
FlG.  14t.  Ans.    -    -   — S  cot   /?;     5  being  the  span 

and /the  inch- stress. 

(a)  If  the  stress  in  BD  or  EC  is  equal  to  the  stress  in  DE,  show  that 

ft  =  60° ;  a  being  the  angle  ABC. 

(b)  The  trusses  are  12  ft.  centre  to  centre ;  the  span  is  40  ft.;  the  hori- 
zontal tie  is  1 6  ft.  long ;  the  rafters  are  inclined  at  60°  to  the  vertical ;  the 
dead  weight  of  the  roof,  including  snow,  is  estimated  at  lolbs.  per  sq.  ft. 
of   roof-surface.      Determine  the  stress  in  each  member  when  a  wind 
blows  on  one  side  with  a  force  of  30  Ibs.  per  sq.  ft.  normal  to  the  roof- 
surface,  assuming  that  the  horizontal  reaction  is  equally  divided  between 
the  supports. 

Ans.  Stress  in  AB  =  8956.8  Ibs. ;  BD  =  10015.2  Ibs.  =  EC\ 
AD  =  2503.8  Ibs.  =  AE\  DE  =  8196  Ibs.; 
AC  =11356.8  Ibs. 

70.  In  the  truss  represented  by  the  accompanying  figure,  the  load 
upon  AB  =  W\ ,  upon  AC  =  W* ;  the  angle  ABD= 

ft\  the  span  BC  —  5;  the  ties  AD,  BD,  AE,  CE  are 
equal ;  /''and  G  are  the  middle  points  of  the  rafters. 
Ftod  the  amount  of  metal  in  the  tie-rods  (wrought- 
iron). 


6  /  sin  ft   cos  ft 

(a)  The  struts  DF  and  EG  are  each  5  ft.;  the  angle  ABC  —  30° ;  the 
dead  weight  of  the  roof,  including  snow,  is  9  Ibs.  per  square  foot  of  roof- 
surface,  and  the  trusses  are  12  ft.  centre  to  centre.  Determine  the 
stresses  in  the  several  members  when  a  wind  blows  with  a  force  of  30 


EXAMPLES.  / 

Ibs.  per  square  foot  of  roof-surface  normal  to  the  side  AB.    The  span 
=  60  ft.,  and  the  end  C  rests  upon  rollers. 

Secondly,  determine  the  stresses  produced  in  the  members  of  the 
truss  in  the  preceding  question  when  a  single  weight  of  3000  Ibs.  is  sus- 
pended from  G. 

Ans.—(\)  Stresses  in       BD ;          DA ;         DE;        EA;        EC; 

31238.55;  19852.35;  12633.6;  8113.5;  24379.43; 
BF;          FA;         FD;  CG;  GA ;          GE. 

29620.44;  28685.16;  7855.2;   22420.44;   21485.16;  1620  ibs. 
(2)  Stresses  in   BD;        DA;          DE;          EA;  EC; 

375  t/39;  125  4/39;  looo  Vj;  875  \/&\  11251/3^; 
BF;        FA;        FD ;        CG;          GA;  GE.  _ 

2625;        2625;          o;  7875;        6375;         1 500  4/3  Ibs. 

(b)  The  rafters  AB,  AC  are  of  unequal  length  and  make  angles  of 
60°  and  45°,  respectively,  with  the  vertical;  the  strut  DF  =  -]\  ft.;  the 
tie  DE  is  horizontal  ;  the  dead  load  upon  each  rafter  =  100  Ibs.  per 
lineal  foot;  the  wind-pressure*normal  to  AB  =  300  Ibs.  per  lineal  foot; 
rollers  are  placed  at  C.  Find  the  stresses  in  all  the  members.  The 
rafter  AB  =  45  ft. 

Show  by  dotted  lines  how  the  stress  diagram  will  be  modified: 

(1)  If  the  rollers  aie  placed  at  B. 

(2)  If  the  strut  DF  is  omitted. 

(3)  If  a  single  weight  of  500  Ibs.  is  concentrated  at  D. 

(V)  If  it  is  assumed  that  the  horizontal  reaction  is  equally  divided  be- 
tween B  and  C,  show  that  the  stress  in  DE  due  to  a  horizontal  wind- 
pressure  upon  AB  is  nil ;  the  angle  ABC  being  30°. 

(d)  In  a  given  roof,  the  rafters  are  of  pitch-pine,  the  tie-rods  of 
wrought-iron  ;  the  span  is  60  ft.;  the  trusses  are  12  ft.  centre  to  centre; 
DF '  =  5  ft. =EG;  the  angle  ABC =  30°;  the  dead  weight  of  the  roof,  in- 
cluding snow,  is  9  Ibs.  per  sq.  ft.  of  roof-surface ;  rollers  are  placed  at  C; 
a  single  weight  of  3000  Ibs.  is  suspended  from  F,  and  the  roof  is  also 
designed  to  resist  a  normal  wind-pressure  of  26.4  Ibs.  per  sq.  ft.  of  roof- 
surface  on  one  side  AB.  Determine  the  stresses  in  the  several 
members. 

71.   In  the  truss  represented  in  the  accompanying  figure,  the  struts 
DF,  DH,  EG,  EK  are  equal,  and  the  ties  BD,  AD, 
EA,  EC  are  also  equal ;  the  load  upon  AB  is  W\ , 
and  upon  AC  is  W* .      Find  the  weight   of  metal 
(wrought-iron)  in  the  ties.  FlG 

Ans     -5-   -    4^i  +  3W  +  ^2)cos2/? 
18  /  cos  ft   sin  ft  • 

(a)  AD  =  AE  =  BD  =  EC  =  23  ft.;  the  angle  ABC  =  30° ;  the  span 
=  79  ft.;  the  trusses  are  13  ft.  centre  to  centre;  the  heel  B  is  free  to 


THEORY  OF  STRUCTURES. 


slide  on  a  smooth  wall-plate ;  the  dead  weight  of  the  roof,  including 
snow,  is  8  Ibs.  per  square  foot  of  roof-surface.  Determine  the  stress  to 
which  each  member  is  subjected  when  the  wind  blows  horizontally  with 
a  force  of  40  Ibs.  per  square  foot  of  vertical  surface  (i)  upon  the  side 
AB,  (2)  upon  the  side  AC. 

Ans.  See  Ex.  3,  Art.  24. 

(b)  The  rafters  AB,  AC  are  inclined  at  60°  to  the  vertical  and  are 

each  40  ft.  in  length.  The  foot  Crests 
on  rollers,  and  the  foot  B  is  fixed.  The 
strut  DF  is  vertical,  is  10  ft.  long,  and 
is  equal  to  the  strut  DE  in  length. 
Also  AF  =  HF=io  ft.  The  dead  load 
carried  by  the  rafters  is  120  Ibs.  per 
lineal  foot.  Provision  has  also  to  be 
made  for  a  normal  wind-pressure  upon 
AB  of  300  Ibs.  per  lineal  foot.  Draw 
the  stress  diagram,  and  show  how  it 
will  be'modified  if  the  strut  DF  is  re- 
FlG-  I44'  moved. 

Ans.  Vertical  reaction  at  B  =  10528  Ibs.  both  before  and  after 

DF  is  removed. 

Horizontal  reaction  at  B~=  6000  Ibs. 
The  dotted  lines  show  the  modified  stresses  for  one  half 
of  the  truss. 

72.  The  load  upon  a  roof-truss  of  the  accompanying  type  is  1000  Ibs.  at 
each  joint;   the  span  100  ft.;   the  rise  =  25  ft.     Find  the  stresses   in 

A 


C 
G  K  N  0  P 

FIG.  145. 

the  different  members.  How  will  the  stresses  be  affected  by  an  addi- 
tional load  of  250  Ibs.  at  each  of  the  joints  between  the  foot  and  ridge 
on  one  side? 

Ans.  Stress  in  BD  =  5500  4/5~;  DF  =  5000  1/5"; 

'  FH '=  4500  4/5  ;  HL  —  4000-4/5';  LN '=  3500  4/5"; 
NA  =  3000  1/5  ;  DE  —  o  ;  FG  =  500 ;  HK  =  1000; 
LM,  =  1 500 ;  NO  =  2000  ;  AP  =  5000  ; 
BE  =  i  looo  =  EG-  GK—  10000  ;  KM  ~  9000  ; 
MO  =  8000 ;  OP  =  7000 ;  DG  =  500  4/5"; 
FK  =  1000  4/2  ;  MM =  500  4/FJ ;  LO  —  1000  4/5"; 
NP  =  500  4/29  Ibs. 


EXAMPLES.  89 

73.  The  dead  load  upon  a  roof-truss  of  accompanying  type  consists 

of  looo  Ibs.  at  F,  1000  Ibs.  at  K,  and  500  Ibs. 
a^  G',  the  wind-pressure  is  a  normal  force  of 
3°  lbs-  Per  scluare  f°ot  of  roof-surface  upon 

^;  the  span  =  9°  ft>;  the  rise  =  2S  ft-;  the 

trusses  are  25  ft.  centre  to  centre.     Find  the 
FlG-  14<5t  stresses  in  the  several  members  when  rollers 

are  (a)  at  C,  (ff)  at  B. 

Ans.  —  (a)  Reaction  (vertical)  at  C=  12291$  Ibs.;  vertical  reaction 
at  ^  =  23958^  Ibs.;  horizontal  reaction  at  B  = 
18750  Ibs. 

Tension  in  BD  —  48625  ;  DL  —  34475  ;  LE  =  21675  '• 
EC=  22125;  Z?//=786i£;  y2£  =  15888!  ; 
A^  =  250  Ibs. 

Compression  in  BF=^666^  4/106  ;  7<y/=2788f  4/106  ; 
/f,4  =  1977$  4/106  ;  ^A"  =  2325  4/106  ; 
ATS  =  2408^  4/106  ;  £C  =  2458!  4/106  ; 

=  1572!  4/106  ;  Z/f=  1505!  4/181  ; 

=  83^  4/T8i  ;  ^6^  =  50  4/7^6  Ibs. 

(<5)  Only  alteration  in  stresses  is  that  each  stress  in  the 
different  sections  of  the  horizontal  tie  is  diminished 
by  18750  Ibs.;  all  the  remaining  stresses  are  un- 
changed. 

74.  In  the  accompanying  roof-truss,  angle  ABC  =  30°  ;  the  span  =  90$- 

ft.;  DF  =  EG  =  io|  ft.;  each  rafter  is  divided  into 
four  equal  segments  by  the  points  of  support  ; 
the  trusses  are  20  ft.  centre  to  centre  ;  the  weight 
of  a  bay  of  the  roof  =  24416  Ibs.  Determine  the 
FIG-  147-  stress  in  each  member. 

Also  determine  the  stresses  due  to  a  wind-pressure  of  30  Ibs.  per  square 
foot  of  roof-surface  acting  normally  to  ABt  when  rollers  are  under  (a)  C, 


75.  The  figure  represents  a  bowstring  truss  of  80  ft.  span,  cut  off  by 
the  plane  MN  and  supported  at  O.    The  upper 
flange  OCDE  is  an  arc  of  a  circle  of  85  ft.  radius  ;  M 

OA  =  AB  =  etc.  =  10  ft.  ;  the  rise  of  the  truss 


=  10  ft.  ;  a  load  of  15  tons  is  concentrated  at  each 

of  the  points  A  and  B  ;  the  reaction  at  O  —  45       " 

tons.     Find  the  stresses  in  the  members  cut  by  |N 

the  plane  MN.  FlG'  I48' 


THEORY  OF  STRUCTURES. 


76.  The  figure  is  a  portion  of  a  bridge-truss  cut  off  by  the  plane  MN 
and  supported   upon   the   abutment   at  A;  AC 
=CE=  14/4  ft.;  the  depth  BC  =  DE  =  i;i  ft.  ; 
in  the  third  panel  the  compression  in  the  upper 
=64,600  ibs.  chord    is   64,600  Ibs.  ;  the  tension  in  the  lower 
chord  is  53,800  Ibs.     Find  the  reaction  at  A,  the 
...,80011)8.     equal  weights  supported  at  C  and  E,  and  the 

diagonal  stress  T. 
N  Ans.-    Reaction  =  19,474  Ibs.;  weight   at  C 


FIG.  149. 


and  at  E  =  9737  Ibs.  ;  T=  17.977 


77.  The  top  beam  of  a  roof  for  a  clear  span  of  96  ft.  consists  of  six 
bars  AB,  BC,  CD,  DE,  EF,  FG,  equal  in  length  and   so  placed    that 
A,  B,  C,  D,  E,  F,  G  are  on  circle  of  80  ft.  radius  ;  the  lower  boom  also 
consists  of  six  equal  rods  AH,  HK,  KL,  LM,  MN,  NG,  the  points  H,  K, 
L,  M,  and  N  being  on  a  circle  of   148  ft.  radius;  B  is  connected  with 
//",  C  with  K,  D  with  L,  E  with  M,  and  F  with  N\  the  opposite  corners 
of  the  bays  are  connected  by  cross-braces  ;  the  end  A  is  fixed  to  its  sup- 
port, G  being  allowed  to  slide  freely  over  a  smooth  bed-plate.    Determine 
graphically  the  stresses  in  the  various  members  when  there  is  a  normal 
wind-pressure  per  lineal  foot  of  460  Ibs.  upon  AB,  340  Ibs.  upon  BC,  and 
60  Ibs.  upon  CD. 

78.  A  bowstring  roof-truss,  with  vertical  and  diagonal  bracing,  of 
50  ft.  rise,  and  five  panels,  is  to  be  designed  to  resist  a  wind  blowing 
horizontally  with  a  pressure  of  40  Ibs.  per  square  foot.     The  depth  of  the 
truss  at  the  centre  is  10  ft.     Determine,  graphically,  the  stresses  in  the 
several  members  of  the  truss,  assuming  that  the  roof  rests  on  rollers 
at  the  windward  support. 

79.  Determine  the  chord,  vertical  and  diagonal  stresses  in  a  Howe 
truss  of  80  ft.  span,  8  ft.  depth,  and  ten  panels,  due  to  a  load  of  40  tons 
(a)  concentrated  at  the  centre  ;  (ff)  concentrated  at  the  third  panel  point  ; 
(c)  uniformly  distributed  ;  (d)  distributed  so  that  5  tons  is  at  first  panel 
point,  10  tons  at  second,  and  25  tons  at  third. 

Ans.   Panel  stresses  in  tension  chord  : 


ISt 

2d 

3d 

4th 

5th 

6th 

7th 

8th 

gth 

loth 

a 

20 

40 

60 

80 

100 

100 

80 

60 

40 

20 

b 

28 

S6 

84 

72 

60 

60 

48 

36 

24 

12 

c 

18 

32 

42 

48 

50 

50 

48 

42 

32 

18 

d 

30 

55 

70 

60 

50 

50 

40 

30 

20 

10 

EXAMPLES. 
Panel  stresses  in  compression  chord  : 


a 

20 

40 

60 

80 

80 

60 

40 

20 

b 

28 

56 

84 

72 

48 

36 

24 

12 

c 

18 

32 

42 

48 

48 

42 

32 

18 

d 

30 

55 

70 

60 

40 

30 

20 

10 

Stresses  in  verticals  : 


a 

2O 

20 

20 

20 

40 

20 

20 

20 

20 

b 

28 

28 

28 

12 

12 

12 

12 

12 

12 

c 

18 

14 

10 

6 

4 

6 

10 

14 

18 

d 

30 

25 

15 

10 

10 

10 

10 

IO 

10 

Diagonal  stresses : 


a 

20 

4/~2 

tons 

in 

each 

diago 

nal. 

b 

284/2 

284/2 

284/2 

124/2 

124/2 

124/2 

124/2 

124/2 

124/2 

124/2 

c 

184/2 

144/2 

104/2 

64/2 

24/2 

2  4/2 

64/2 

104/2 

144/2 

l8\/2 

d 

304/2 

254/2 

154/2 

104/2 

104/2 

104/2 

IO|/2 

104/2 

104/2 

104/2 

4/3; 
4/3 


80.  A   Warren   girder  of  60   ft.  span,  composed   of  six   equilateral 
triangles,  carries  upon  its  lower  chord  a  weight  of  2  tons  at  the  first  and 
second  joints,  15  tons  at  the  centre  joint,  and  Ji  tons  at  the  fourth  and 
fifth  joints.     Find  the  stresses  in  all  the  members. 

Ans.  Stresses  in  tension  chord  :  ist  bay  == 

3d  = 
5th  = 

Stresses  in  compr.  chord  :   ist  bay  = 

3d  = 
5th  =_V  ift 

Diag.  stresses  ist  and  2d  bays  =  -1/  4/3  »'  3d  and  4th  =  *$-  4/3"; 

5th  and  6th  =  V-  i/3";    ;thand    8th  =  -1/  4/3"; 

9th  and  loth  =  ^  4/J;  nth  and  I2th  =  ^-  4/3"; 

81.  Determine  the  stresses  in  the  members  of  a  Fink  truss  of  240  ft. 
span  and  sixteen  panels;  depth  of  truss  =  30  ft.;   uniformly  distributed 
load  =  W. 


-  4/3";    2d  = 
-/-  4/3";  4th  = 

-  Vl>\  6th  =  ||  ^J. 
-v/J";    2d  =  %3-  4/3"; 

£  4/3    4th  =  S£  Vy, 


92 


THEORY  OF  STRUCTURES 


Ans.—  Ai    LMNOPQRS'     Stress  in  BA,  BM,  DM,  DO,  FO,  FQ, 

HQ,  H$,  same   and  = 


C    D    E    F    G   H 
FIG.  150. 


64 


in 


w  w    _ 

in  EA,  ES  same  and  =  -=-  |/c  ;  in  AK  =  —  y  17  ;   i 
5  4 

W^  W 

DN,  FP,  HR,  same  and  =  —  r  ;  in  O/,  (7(9  same  and  =  -5-  ; 

10  o 

W  W 

mEO  =  —  ;  in  KS  =  —  ;  in  AM,  MO,  OQ,  QS  same  and 


64 

82.  Determine  the  stresses  in  the  members  of  a  Bollman  truss  100  ft. 
long  and  12^  ft.  deep,  under  a  uniformly  distributed  load  of  200  tons,  to- 
gether with  a  single  load  of  10  tons  concentrated  at  25  ft.  from  one  end. 


Ans.  Stress   in  AB  =  ^  ^2 ;  BL  =  if*  4;  ^Z?  =  y  4/5  ; 
,DZ  =  -2/  4/37  ;  AF  =  H1  I/To;  FL  =  y  1/26; 
^^  ^  3^  ^jy  =///:;       in   ^C  =  25  =  /re  = 
HK  =  etc.;  DE  =  50  tons;   compression  along 
^4£  =  193!  tons. 

Note. — Questions  53,  54,  57-59,  61,  66,  67,  70,  71,  73,  and  74  can  be 
easily  solved  graphically. 

83.  Determine  the  stresses  in  the  several  members  when  the  throw 
of  the  crane  in  Question  55  is  increased  by  the  introduction  of  the  new 
members,  shown  by  the  dotted  lines. 


CHAPTER   II. 
SHEARING   FORCES   AND   BENDING   MOMENTS. 

Note. — In  this  chapter  it  is  assumed  that  all  forces  act  in  one  and  the  same 
plane,  and  that  the  deformations  are  so  small  as  to  make  no  sensible  alteration 
either  in  the  forces  or  in  their  relative  positions. 

I.  Equilibrium  of  Beams. — A  beam  is  a  bar  of  somewhat 
considerable  scantling,  supported  at  two  points  and  acted  upon 
by  forces  perpendicular  or  oblique  to  the  direction  of  its  length. 

CASE  I.  AB  is  a  beam  resting  upon  two  supports  in  the 
same  horizontal  plane.  The  reactions  Ro 

Rl  and  R^  at  the  points  of  support  are 
vertical,  and  the  resultant  P  of  the 
remaining  external  forces  must  also 
act  vertically  in  an  opposite  direction 
at  some  point  C. 


IA 


FIG.  152. 

According  to  the  principle  of  the  lever, 


PAC 
=  P  -r-f* ,     and 


CASE  II.     AB  is  a  beam  supported  or  fixed  at  one  end. 
Such   a   support   tends   to    prevent    any  deviation    from    the 
straight  in   that  portion  of   the  beam, 
and    the  less    the    deviation    the  more 
'B  perfect  is  the  fixture. 

The  ends  may  be  fixed  by  means  of 
two  props  (Fig.  153),  or  by  allowing  it 
,B  to  rest  upon  one  prop  and  preventing 
upward  motion  by  a  ledge  (Fig.  154),  or 
by  building  it  into  a  wall  (Fig.  155). 

In  any  case  it  may  be  assumed  that 
the  effect  of  the  fixture,  whether  perfect 
or  imperfect,  is  to  develop  two  unequal 
forces,  Q  and  R,  acting  in  opposite  di- 


FIG.  155. 


rections  at  points  M  and  N. 


These  two  forces  are  equivalent 

93 


94 


THEORY  OF  STRUCTURES. 


to  a  left-handed  couple  (Q,  —  Q),  the  moment  of  which  is 
Q.MN,  and  to  a  single  force  R—  Q  at  N.  Hence  R  —  Q 
must  =  P. 

CASE  III.  ^4/?  is  an  inclined  beam  supported  at  A  and 
resting  upon  a  smooth  vertical  surface 
at  B. 

The  vertical  weight  P,  acting  at  the 
point  C,  is  the  resultant  load  upon  AB. 
Let  the  direction  of  P  meet  the  hori- 
zontal  line  of  reaction  at  B  in  the 
point  D. 

The  beam  is  kept  in  equilibrium  by 
the  weight  P,  the  reaction  R^  at  A,  and  the  reaction  7v?2  at  B. 
Now  the  two  forces  R^  and  P  meet  at  /),  so  that  the  force  R^ 
must  also  pass  through  D. 


Hence 


cos 


and 


^  =  P  tan 


.  —  The  same  principles  hold  if  the  beam  in  Cases  I  and 
II  is  inclined,  and  also  whatever  may  be  the  directions  of  the 
forces  P  and  R^  in  Case  III. 

CASE  IV.  hi  general,  let  the  beam  AB  be  in  equilibrium 
under  the  action  of  any  number  of  forces  Plt  P2  ,  P9  ,  .  .  .  , 
Q\  *  Qi  >  Qs  >  •  •  •  >  °f  which  the  magnitudes  and  points  of  appli- 


?  & 

f 

M 

?' 

r 

1 

! 

1^ 
FIG. 

157- 

v     v 

Q3  Q2 

cation  are  given,  and  which  act  at  right  angles  to  the  length  of 
the  beam.  Suppose  the  beam  to  be  divided  into  two  segments 
by  an  imaginary  plane  MN.  Since  the  whole  beam  is  in  equi- 
librium, each  of  the  segments  must  also  be  in  equilibrium. 
Consider  the  segment  AMN. 


EQUILIBRIUM   OF  BEAMS.  95 

It  is  kept  in  equilibrium  by  the  forces  Plt  P9,  P9,  .  .  .  and 
by  the  reaction  of  the  segment  BMN  upon  the  segment  AMN 
at  the  plane  MN;  call  this  reaction^.  The  forces  P, ,  P, , 
P3 ,  .  .  .  are  equivalent  to  a  single  resultant  R1  acting  at  a  point 
distant  rl  from  MN.  Also,  without  affecting  the  equilibrium, 
two  forces,  each  equal  and  parallel  to  Rt ,  but  opposite  to  one 
another  in  direction,  may  be  applied  to  the  segment  AMN  at 
the  plane  MN,  and  the  three  equal  forces  are  then  equivalent 
to  a  single  force  R,  at  MN,  and  a  couple  (Rl ,  —  R,)  of  which 
the  moment  is  Rlrl. 


i  *       !«         rR' 
A.       I         !11         ' 

FIG.  158. 

Thus  the  external  forces  upon  AMN  are  reducible  to  a 
single  force  R1  at  MN,  and  a  couple  (Rlt  —  RJ.  These  must 
be  balanced  by  E^ ,  and  therefore  £l  is  equivalent  to  a  single 
force  —  Rt  at  MN  and  a  couple  (—  R^ ,  R^. 

In  the  same  manner  the  external  forces  upon  the  segment 
BMN  are  reducible  to  a  single  force  R^  at  MN,  and  a  couple 
(/?,,  —  ^2)  of  which  the  moment  is  R^.  These  again  must 
be  balanced  by  £3,  the  reaction  of  the  segment  ^4 J£/V  upon 
the  segment  BMN. 

Now  El  and  £2  evidently  neutralize  each  other,  so  that  the 
force  R^  and  the  couple  (R1 ,  —  R})  must  neutralize  the  force 
^2  and  the  couple  (R9,  —  R^).  Hence  the  forced,  and  the 
couple  (Rt ,  —  R^  are  respectively  equal  but  opposite  in  effect 
to  the  force  ^2  and  the  couple  (Rt,  —  R9) ;  i.e., 

Rl  ==  RZ     and     R1rl  =  R^  ;     /.  rl  =  r.t. 

The  force  R,  tends  to  make  the  segment  AMN  slide  over 
the  segment  BMN  at  the  plane  MN,  and  is  called  the  Shearing 


THEORY  OF   STRUCTURES. 


Force  with  respect  to  that  plane.     It  is  equal  to  the  algebraic 
sum  of  the  forces  on  the  left  of  MN, 


So  ^2  =  0,  —  Q,  —  Q3  +  .  .  .  =  2(Q)  is  the  algebraic  sum 
of  the  forces  on  the  right  of  MN,  and  is  the  force  which  tends 
to  make  the  segment  BMN  slide  over  the  segment  AMN  at 
the  plane  MN.  R^  is  therefore  the  Shearing  Force  with  respect 
to  MN,  and  is  equal  to  Rl  in  magnitude,  but  acts  in  an  opposite 
direction. 

Again,  let/,  ,  /2  ,  /3  ,  .  .  .  ,  ^  ,  qz  ,  qz  ,  .  .  .  ,  be  respectively  the 
distances  of  the  points  of  application  of  Pl  ,  Pt  ,  P3  ,  .  .  .  ,  Ql  ,  <22  » 
<2,  ,  .  .  .  from  MN. 

Then  R1r1  ,  =  the  algebraic  sum  of  the  moments  about 
MN  of  all  the  forces  on  the  left  of  MN, 


is  the  moment  of  the  couple  (Rlt  —  R^). 

This  couple  tends  to  bend  the  beam  at  the  plane  MN,  and 
its  moment  is  called  the  Bending  Moment  with  respect  to  MN 
of  all  the  forces  on  the  left  of  MN. 

So  R9r9  ,  =  the  algebraic  sum  of  the  moments  about  MN 
of  all  the  forces  on  the  right  of 


is  the  Bending  Moment,  with  respect  to  MN,  of  all  the  forces  on 
the  rig/it  of  MN,  and  is  equal  but  opposite  in  effect  to  Rlr1. 

It  is  seen  that  the  Shearing  Force  and  Bending  Moment 
change  sign  on  passing  from  one  side  of  MN  to  the  other,  so 
that  to  define  them  absolutely  it  is  necessary  to  specify  the  seg- 
ment under  consideration. 

Remark.  —  The  reaction  El  has  been  shown  to  be  equivalent 
to  the  force  —  Rl  and  the  couple  (—  Rl  ,  R^.  The  Moment 
of  this  couple  may  be  called  the  Elastic  Moment,  the  Moment 
of  Resistance,  or  the  Moment  of  Inflexibility,  and  is  equal  in 
magnitude,  but  opposite  in  effect,  to  the  corresponding  Bend- 
ing Moment  due  to  the  external  forces. 


SHEARING  FORCES  AND  BENDING  MOMENTS. 


97 


FIG.  159. 


2.  Examples  of  Shearing  Forces   and  Bending  Mo- 
ments.— In    each    of   the   following 
examples  the  beam  is  horizontal  and 
of  length  /.  . 

Ex.  i.  The  beam  OA,  Fig.  159, 
is  fixed  at  A  and  carries  a  weight 
Pat  O. 

The  Shearing  Force  (S)  at  every 
point  of  the  beam  is  evidently  con- 
stant and  equal  to  P. 

Upon  the  verticals  through  A  and  0  take  AB  and  OC  each 
equal  or  proportional  to  P;  join  BC.  The  vertical  distance 
between  any  point  of  the  beam  and  the  line  BC  represents  the 
shearing  force  at  that  point. 

Again,  the  Bending  Moment  (M)  at  any  point  of  the  beam, 
distant  x  from  O  is  Px ;  it  is  nil  at  <9,  and  PI  at  A. 

Upon  the  vertical  through  A  take  AD  equal  or  propor- 
tional to  PI;  join  DO.  The  vertical  distance  between  any 
point  of  the  beam  and  the  line  DO  represents  the  bending 
moment  at  that  point. 

Ex.  2.  The  beam  OA,  Fig.  160,  is  fixed  at  A,  and  carries 
a  uniformly  distributed  load,  of  in- 
tensity w  per  unit  of  length. 

The  resultant  force  on  the  right 
of  a  vertical  plane  MN  distant  x  from 
O  is  wx  and  acts  half-way  between 
)   O  and  MN. 

The  Shearing  Force  (S)  at  MN  is 
therefore  wx ;  it  is  nil  at  (9,  and  wl 
at  A.  Upon  the  vertical  through  A 
take  AB  equal  or  proportional  to  wl\  join  BO.  The  vertical 
distance  between  any  point  of  the  beam  and  the  line  BO  rep- 
resents the  shearing  force  at  that  point. 


FIG.  160. 


Again,  the  Bending  Moment  (M)  at  MN  is  ivx—  = 


wx 


it 


is  nil  at 


O,  and at  A. 


Upon  the  vertical  through  A  take 


AC  equal  or  proportional  to . 


98 


THEORY  OF  STRUCTURES. 


The  bending  moment  at  any  point  of  the  beam  is  repre- 
sented by  the  vertical  distance  between  that  point  and  a  pa- 
rabola CO  having  its  vertex  at  O  and  its  axis  vertical. 

Ex.  3.     The  beam  OA,  Fig.  161,  is  fixed  at  A  and  carries 


wx 


FIG.  161. 

a  single  weight  P  at  O,  together  with  a  uniformly  distributed 
load  of  intensity  w  per  unit  of  length. 

The  Shearing  Force  (S)  at  a  plane  MN  distant  x  from  O 
is  evidently  P '  +  wx  \  it  is  P  at  (9,  and  P-\-  wl  at  A. 

Upon  the  verticals  through  O  and  A  take  OC  equal  or  pro- 
portional to  P,  and  AB  equal  or  proportional  to  wl -\- P\ 
join  BC.  The  vertical  distance  between  any  point  of  the 
beam  and  the  line  BC  represents  the  shearing  force  at  that 
point. 

Again,  the  Bending  Moment  (M)  at  MN  is  evidently 


it  is  nil  at  O,  and  ----  \-  PI  at  A. 

Upon   the  vertical  through  A  take  AD  equal  or  propor- 


tional   to  --  1-  PL     The  bending  moment   at   any  point  of 

the  beam  is  represented  by  the  vertical  distance  between  that 
point  and  a  parabola  DOE  having  its  axis  EF  vertical  and  its 


SHEARING  FORCES  AND  BENDING  MOMENTS.  99 

p 

vertex  at  a  point  E,  where  OF  =  —  and  EF  is  equal  or  pro- 

portional  to  — . 
2w 

Note. — The  ordinates  of  the  line  BC  in  Ex.  3,  are  equal  to 
he  algebraic  sum  of  the  corresponding  ordinates  of  the  straight 
lines  BC  and  BO  in  Exs.  I  and  2.  Also,  the  ordinates  of  the 
curve  DO  in  Ex.  3,  are  equal  to  the  algebraic  sum  of  the  cor- 
responding ordinates  of  the  line  DO  in  Ex.  I,  and  the  curve 
CO  in  Ex.  2.  Hence  the  same  conclusions  as  in  Ex.  3  are 
arrived  at  by  treating  the  weight  P  and  the  load  wl  inde- 
pendently, and  then  superposing  the  respective  results. 

Ex.  4.  The  beam  OA,  Fig.  162,  rests  upon  two  supports 
at  O  and  A,  and  carries  a  weight  P 
at  a  point  B,  dividing  the  beam 
into  the  two  segments  OB,  J3A,  of 
which  the  lengths  are  a  and  b  re- 
spectively. 

The  reactions  Rl ,  Rz  at  O  and 
A  are  vertical,  and  according  to  the 
principle  of  the  lever,  'FIG. 

R,  —  P~t     and     R,  =  P*. 

The  Shearing  Force  (S)  at  every  point  between  O  and  B  is 
constant  and  equal  to  Rl  =  P-,.  On  passing  B  the  shearing 
force  (S)  changes  sign,  and  its  value  at  every  point  between  B 
-ind  A  is  constant  and  equal  to  R^  —  P  =  —  P-,  —  —  R^ .  Upon 
the  verticals  through  O,  B,  and  A  take  OC\  BE,  each  equal  or 
proportional  to  -y,  and  BF,  AD,  each  equal  or  proportional 

to  -y  ;  join  CE  and  DF.     The  shearing  force  at  any  point  of 

the  beam  is  represented  by  the  vertical  distance  between  that 
point  and  the  broken  line  CEFD. 


100  THEORY   OF  STRUCTURES. 

Again,  the  Bending  Moment  (M)  at  any  point  between   O 
and  B  distant  x  from  O  is  Rlx=^P-.x-J   it  is  nil  at  O,  and 

P^j  at  B. 

The  Bending  Moment  (M)  at  any  point  between  B  and  A 

distant    x    from    O    is   R^x  —  /%*;  —  a)  =  P-.(l  —  x)  ;    it    is 

ab 
P-j  at  B,  and  nil  at  A. 

Upon  the  vertical  through  B  take  BG  equal  or  proportional 

ab 
to  P~r\   join   06-    and   y^^.     The    bending    moment    at    any 

point  of  the  beam  is  represented  by  the  vertical  distance  be- 
tween that  point  and  the  line  OGA. 

p 
Cor. — If  Pbe  at  the  centre  of  the  beam,  S  =  —,  and  M 

PI 

at  the  centre  =  — . 

4 

Ex.  5.     The  beam  OA,  Fig.  163,  rests  upon  two  supports 

at  O  and  A,  and  carries  a  uni- 
formly distributed  load  of  inten- 
sity w  per  unit  of  length. 

The  reactions  at  O  and  A  are 

each  equal  to  — . 

The  resultant  force  between  O 
and  a  plane  MN  distant  x  from  O 
FlG- 1<53-  is  wxy  and  acts  half-way  between 

O   and  MN.     The    Shearing  Force  (S)  at  MN  is   theiefoie 

wl  wl 

wx ;    it    is  at   O,   nil    at    the    middle    point   B,  and 

2  2 

-  at  A.     Upon   the  verticals  through   O  and  A   take  OC 

and  ADj  each   equal   or   proportional   to  --  ;  join   CD.     The 

shearing  force  at  any  point  of  the  beam  is  represented  by  the 
vertical  distance  between  that  point  and  the  line  CD. 


SHEARING   FORCES  AND   BENDING   MOMENTS. 

Again,  the  Bending  Moment  (M)  at  MN  is 


101 


wl 


wl 


wx" 

., 


it  is  nil  at  O  and  at  A  ;  it  is  a  maximum  and  equal  to  -—  at 

b 

the  middle  point  B.  Upon  the  vertical  through  B  take  BE 
equal  or  proportional  to  -— .  The  bending  moment  at  any 

o 

point  of  the  beam  is  represented  by  the  vertical  distance  be- 
tween that  point  and  a  parabola  OEA  having  its  vertex  at  E 
and  its  axis  vertical. 

Cor.  i.  The  shearing  force  is  a  minimum  and  zero  at  the 

Wl 

centre,  a  maximum  and  —  at  the  ends,  and  increases  uni- 
formly with  the  distance  from  the  centre. 

Cor.  2.  The  bending  moment  is  a  minimum   and   zero  at 

w/2 

the  ends,  a  maximum  and  -=-  at  the  centre,  and  diminishes 

o 

as  the  distance  from  the  centre  increases. 

Ex.  6.  The  beam  OA,  Fig.  164,  rests  upon  two  supports  at 
O  and  A,  and  carries  a  weight  P  at 
a  point  B,  together  with  a  uniformly 
distributed  load  of  intensity  w  per 
unit  of  length. 

Let  the  lengths  of  the  segments 
OB,  BA  be  a  and  b,  respectively. 

The  reactions  R^  at  O,  and  R^ 
at  A,  are  vertical,  and  according  to 
the  principle  of  the  lever, 


b      wl 


and 


FIG.  164. 


IO2         .  THEORY  OF  STRUCTURES. 

The  Shearing  Force  (S)  at  any  vertical  plane  between  O  and 
B  distant  x  from  O  is 

b       wl 
Rl  —  vox  =  P-j  -\ wx ; 

/  2 

Pb       wl                         Pb    ,   w/ 
it  is     -T-  -| -at  (9,     and     —j-  -| w#  at  /?. 

/  2  /  2 

The  Shearing  Force  (S)  at  any  plane  between  ^  and  ^4 
distant  ;r  from  6>  is 


R^-P-wx^P-f  +  ^-P-wx^^-P-j-wx-, 

wl       Pa                                        Pa       wl 
it  is j wa  at  B,     and = at  A . 

21  I  2 

Upon  the  verticals  through  O,  B,  and  A  take  OC  equal  or 
proportional  to  — -, — | — —  ,  BD  equal  or  proportional  to 

I  2 

Pb       wl                                                                   wl      Pa 
—j--\- wa,  BE  equal  or  proportional  to -. wa, 

wl       Pa 
and  AF  equal   or   proportional  to -=- ;  join  CD  and 

2  / 

EF.  The  shearing  force  at  any  point  of  the  beam  is  rep- 
resented by  the  vertical  distance  between  that  point  and  the 
broken  line  CDEF. 

wl       Pa 
If  • —  >  -7-  +  wa,  BE  is  positive,  and  therefore  E  is  ver- 

2          •    / 

tically  above  B. 

Again,  the  Bending  Moment  (M)  at   any  point  between 
O  and  B  is 


it  is  nil  at  0,  and 


SHEARING  FORCES  AND   BENDING  MOMEN7^S.          103 

The  bending  moment  (M)  at  any  point  between  B  and  A 
distant  x  from  O  is 

/    b      wl\         wx*  I       _a   ,   wl\         wx* 

(P-I  +  -J)*-  —  -*\*  -*)  =  (-  p-i  +  -?)*—T+pa> 

/    b       wl\         wa* 
it  is  \P-j  +  —In  --  at  £,  and  nil  at  A.     , 

Upon  the  vertical  through  j£  take  BG  equal  or  propor- 

(^       wl\         we? 
P-j-\  --  Jtf  --  .      The    bending    moment    at 

any  point  of  the  beam  between  O  and  B  is  represented  by  the 
vertical  distance  between  that  point  and  a  parabola  OGH 
having  its  axis  HT  vertical  and  its  vertex  at  a  point  //,  where 


and 

HT  is  equal  or  proportional  to  —  \P-.  -|  --  )  . 

2,1.U  S    •/  2   ' 

The  bending  moment  at  any  point  between  B  and  A  is 
represented  by  the  vertical  distance  between  that  point  and  a 
parabola  AGK  having  its  axis  KV  vertical  and  its  vertex  at  a 
point  K,  where 


and 

a       wlY 


KV  is  equal  or  proportional  to  — 1  —  P-.  -| j   -|~ 


Cor, — If  the  weight  Pis  at  the  centre, 

P  PI       wl* 

S  =  — ,  and  M  at  the  centre  = 1-  -5-. 

2  4  o 

Note. — The  ordinates  of  the  lines  CD  and  EF  in  Ex.  6  are 
equal  to  the  algebraic  sum  of  the  corresponding  ordinates  of  the 


104 


THEORY  OF  STRUCTURES. 


lines  CE,  FD  in  Ex.  4,  and  the  line  CD  in  Ex.  5.  Also,  the 
•ordinates  of  the  curves  OGy  AG  are  equal  to  the  algebraic 
sum  of  the  corresponding  ordinates  of  the  lines  OG,  AG  in  Ex. 
4,  and  the  curve  OEA  in  Ex.  5.  Hence  the  same  conclusions 
as  in  Ex,  6  are  arrived  at  by  treating  the  weight  P  and  the 
load  wl  independently,  and  then  superposing  the  respective 
results. 

Ex.  7.  In  fine,  a  beam,  however  loaded,  may  be  similarly 
treated,  remembering  that  if  the  load  changes  abruptly  at  dif- 
ferent points,  the  portions  of  the  beam  between  the  points  of 
discontinuity  are  to  be  dealt  with  separately.  For  example, 
the  beam  OA,  Fig.  165,  rests  upon  two  supports  at  0  and  A, 
and  carries  three  weights  P, ,  P9 ,  P3  at  points  C,  D,  E,  of  which 
the  distances  from  O  are  p^  /3,-/3,  respectively.  A  point  B 
divides  OA  into  segments  OB  =  a  and  BA  =  #,  which  are 


FIG   165. 


uniformly  loaded  with  weights  of  intensities  wv  and  w^  per 
unit  of  length,  respectively.  The  reactions  R}  and  R^  at  O  and 
A  are  vertical,  and  according  to  the  principle  of  the  lever, 


RJ  =  />,(/  -  A)  +  Ptf  -  A}  +  />,(/  -  A) 


a 


and 


+ «X4 + 4 


SHEARING  FORCES  AND  BENDING  MOMENTS.          10$ 

To  represent  graphically  the  Shearing  Force  at  different 
points  of  the  beam  : 

Upon  the  verticals  through  <9,  C,  B,  D,E,  A,  take  OF,  CG, 
CH,  BK,  DL,  DM,  EN,  EV,  and  A  T,  respectively  equal  or 
proportional  to 


R,,  R,  —  wj,,  R  -  wj,  -  />,,  R,  -  w,a  -r-  Plt 
R,  -  w,a  —  P,  —  ?e>2(/2  -  a\ 
R,  -  w,a,  -  Pl  -  w2(A  -  a)  -  P» 
R,  -  w,a  —  Pl  -  wjpt  -  a)  -  P9, 
Rv  -  ^vla  -  Pl  -  u>,(p,  —  a)  -  P,  -  P9, 
and         R,  -  w,a  -  P,  -  w,b  -  P,  -  P,  =  R,. 


Join  FG,  HK,  KL,  MN,  and  VT.  The  shearing  force  at 
any  point  of  the  beam  is  represented  by  the  vertical  distance 
between  that  point  and  the  broken  line  FGHKLMNVT. 

To  represent  graphically  the  Bending  Moment  (M)  at  dif- 
ferent points  of  the  beam  : 


Mat  0  =  0',         Mat  C=Rj>l  -  ; 


Mat  D  =  R.tp,  -  ?ey?(2  - 


106 


THEORY  OF  STRUCTURES. 


-  ^(A_^T  _  PI(A  _  A)  _  />i(A  _  A) . 


and 


MatA=0. 


Upon  the  verticals  through  C,  B,  D,  and  £  take  Ci,  B2, 
Z>3,  and  £4,  respectively  equal  or  proportional  to  the  bending 
moments  at  these  points. 

The  bending  moment  at  any  point  of  the  beam  is  repre- 
sented by  the  vertical  distance  between  that  point  and  the 
parabolic  arcs  Oi,  12,  23,  34,  and  4.A.  The  axes  of  these  pa- 
rabolas are  vertical,  and  the  positions  of  the  vertices  may  be 
easily  found  from  the  several  equations. 

Ex.  8.  A  beam  OA,  Fig.  166,  of  which  the  weight  may  be 

neglected,  is  15  ft.  long,  is  fixed 
at  (9,  and  carries  a  weight  of  80 
Ibs.  at  A.     Determine  the  bend- 
ing moment  at  a  point  distant  10 
ft.  from  the  free  end.     Also  illus- 
trate the  shearing  force  and  bend- 
ing moment  at  different  points  of 
the   beam    graphically.      The   re- 
quired bending  moment  is  80  X 
10  —  800  Ib.-ft. 
The  shearing  force  is  the  same  at  every  point  of  the  beam, 
and  equal  to  80  Ibs.     Choose  a  vertical  scale  of  measurement 
so  that  half  an  inch  represents  160  Ibs. 

Upon  OA  describe  a  rectangle  OABC,  in  which  OC  =  AB 
=  y.  The  ordinate  from  every  point  of  BC  to  AO  is  J",  or 
80  Ibs.,  and  is  therefore  the  shearing  force  at  the  foot  of  such 
ordinate. 

Again,  the  bending  moment  at  O  is  80  X  15  =  1200  Ib.-ft. 
Choose  a  vertical  scale  of  measurement  so  that  I  inch  repre- 
sents 1 200  Ib.-ft.  Upon  the  vertical  through  O  take  OD  = 


FIG.  166. 


SHEARING  FORCES  AND   BENDING  MOMENTS. 


107 


I  inch ;  join  DA.  The  ordinate  from  any  point  of  DA  to  OA 
is  the  bending  moment  at  its  foot.  For  example,  at  ii£  ft. 
from  O  the  ordinate  is  J",  or  300  lb.-ft.,  and  this  is  equal  to 
80  X  3i',  i.e.,  the  bending  moment. 

Ex.  9.  A  beam  OA,  Fig.  167,  of  which  the  weight  may  be 
neglected,  rests  upon  two  supports 
at  0  and  A,  30  ft.  apart,  and  carries 
a  uniformly  distributed  load  of  200 
Ibs.  per  lineal  foot,  together  with  a 
single  weight  of  600  Ibs.  at  a  point 
B  dividing  the  beam  into  segments 
OB,  BA,  of  which  the  lengths  are  10 
and  20  ft.  respectively.  Determine 
the  shearing  force  and  bending  mo- 
ment at  the  points  C  and  D,  distant 
5  ft.  from  the  nearest  end.  Also, 
illustrate  graphically  the  shearing 
force  and  bending  moment  at  differ- 
ent points  of  the  beam. 

Let  Rl ,  R^  be  the  reactions  at  O 
and  A,  respectively.     Then 

R, .  30  —  600 .  20  +  200 .  30 .  1 5  =  102000 ; 

.-.  Rl  =  3400  Ibs.,  and  R^  —  200 .  30  +  600  —  Rl  =  3200  Ibs. 

The  Shearing  Force  at  C  =  3400  —  200 .  5  —  2400  Ibs. 

"  £>=3400-200.  25-600=  —  2200  Ibs. 

The  Bending  Moment  at  C  =  3400  .  5  —  200  .  5  .  - 

=  14,500  lb.-ft. 
The  Bending  Moment  at  D  =  3400. 25  —  200 .  25 .  —  —  600. 1 5 

=  13,500  lb.-ft. 

Next,  considering  the  segment  OB,  the  shearing  force  at  O 
is  3400  Ibs.,  and  at  B  1400  Ibs. 

Considering  the  segment  BA,  the  shearing  force  at  A   is 
—  3200  Ibs.,  and  at  B  800  Ibs. 

Choose  a  vertical  scale  of  measurement  so  that  i  inch  repre- 
sents 3000  Ibs.     Upon  the  verticals  through  O,  B,  A  take  OE 

-  -fa" ,  and  AH  =  iyy  ;  join  EF  and 


FIG.  167. 


IOS  THEORY   OF  STRUCTURES. 

GH.  The  ordinate  from  any  point  of  the  broken  line  EFGH  to 
OA  is  the  Shearing  Force  at  its  foot.  For  example,  the  ordinate 
at  D  is  —  ||",  or  —  2200  Ibs. 

Again,  the  bending  moment  at  B  is  3400.  10—  200.  10.  5 
=  24,000  Ib.-ft.  Choose  a  vertical  scale  of  measurement  so  that 
I  inch  represents  24,000  Ib.-ft.  Upon  the  vertical  through  B 
take  BK  =  i  inch.  Draw  the  parabolas  OK,  AK,  with  their 
vertices  at  points  determined  as  in  Example  (6).  The  ordi- 
nate from  any  point  of  the  curves  OK,  AK  is  the  bending 
moment  at  its  foot. 

For  example,  at  a  point  14  ft.  from  O  the  curve  ordinate  is 
i-jJy",  or  25,600  Ib.-ft.,  and  this  is  the  Bending  Moment  at  the 
same  point,  being  also  the  greatest  for  the  segment  BA.  The 
vertex  of  AK  is,  therefore,  vertically  above  the  point  of 
which  the  horizontal  distance  from  O  is  14  ft. 

3.  Relation  between  Shearing  Force  and  Bending 
Moment. — Let  a  beam  AB  be  arbitrarily  loaded  with  weights 
wl ,  w2 ,  wz ,  .  .  .  concentrated  at  the  points  I,  2,  3,  ... 

A  1,  2,        3  rrl       r       r-+l n__    B 


FIG.  168. 

Let  «, ,  #2 ,  #3 ,  .  .  .  be  the  lengths  of  the  segments  Ai,  12, 
23,  .  .  .  ,  respectively. 

Let  MA  ,  MB  be  the  moments  at  A  and  B.  These  moments 
are  of  course  nil  if  the  beam  merely  rests  upon  supports  at  its 
ends. 

The  reaction  R  at  A  is  given  by  the  equation 

Rl  =  «/,(/-*,)  +  wJJ  -a1-a,)+...+MB  +  MA) 
I  being  the  length  of  the  beam. 

The  shearing  force  S,  between  A  and  i  =  R ; 

3  O  12' 

Sn        "     «-i  "     n  =  R—  2(w)\ 
2(w)  denoting  the  sum  of  the  first  («  —  i)  weights. 


SHEARING  FORCE  AND  BENDING  MOMENT.  1  09 

The  bending  moment 

MA  -atA=MA  ; 

=  M 


,  ,  A          A 

,  "    2  =  ^<X  +  a,)  -  «v*,  +  MA  =  M, 


M 


Hence  the  difference  between  the  bending  moments  at  the  be- 
ginning and  end  of  any  interval  is  equal  to  the  product  of  the 
shearing  force  (S)  for  that  interval  by  the  length  (a)  of  the  in- 
terval. 

Let  AM  denote  the  difference  between  any  two  consecutive 
bending  moments  ;  then 


This  result  has  been  deduced  without  any  assumption  as  to 
the  number  of  the  loads.  They  may  therefore  be  infinite  in 
number  and  in  the  limit  form  a  continuous  load. 

Thus,  if  5  be  the  shearing  force  at  a  distance  x  from  A, 


dM 


or,  the  shearing  force  at  any  point  is  equal  to  the  rate  of  increase 
of  the  bending  moment  per  unit  of  length. 

The  above  results  may  also  be  expressed  as  follows :  The 
shearing  force  at  any  point  is  measured  by  the  tangent  of  the  slope 
at  the  corresponding  point  of  the  b ending-moment  polygon  or 
curve. 

The  shearing  force  is  positive,  zero,  or  negative  according  as 


R  =  2(w) ; 


IIO  THEORY  OF  STRUCTURES. 

2(w)  being  the  sum  of  the  weights  up  to  the  point  under  con- 
sideration.     In  the  case  of  a  continuous  load,  of  intensity  GO, 


-,(w)  =    I  xwdx. 


Thus  the  bending  moment  M  at  the  same  point  is  a  maxi- 
mum (or  a  minimum  in  certain  special  cases)  when  the  shear- 
ing force  changes  sign,  i.e.,  when 


Again,  with  an  arbitrarily  distributed  load 


and  with  a  continuous  load 


Thus  the  difference  between  the  ordinates  of  the  bending- 
moment  diagram  at  any  point  and  A  is  proportional  to  the 
area  of  the  shearing-force  diagram  between  the  same  points. 
From  this  result  an  important  deduction  may  at  once  be 
made. 

The  bending  moment  Mx  at  any  point  between  r  and  r  +  I 
distant  x  from  r  is 


—  Mr  +  x(R  —  wl  —  w2  —  .  .  .  —  wr) 

=  Mr  +  xSr+l. 

Now    Mr+l  =  Mr  -f  ar+lSr+iy         and  therefore  Sr+l  is  zero  if 
Mr+l  =  Mr  ,     and  also    Mx  =  Mr  =  Mr+1  . 

Thus,  the  bending  moment  is  the  same  at  every  point  between 
r  and  r  -f-  i,  and  the  case  is  one  of  simple  bending  without  shear, 
as,  e.g.,  with  a  carriage-axle. 


EFFECT  OF  A    ROLLING   LOAD. 


Ill 


4.  To  Discuss  the  Effect  of  a  Rolling  Load.— CASE  I. 

Let  a  single  weight  W  travel  from  left  to  right  over  a  girder 
OA    of    length  /,  resting    upon   two 
supports  at  O  and  A. 

The  reaction  Rl  at  (9,  when  W  is 

at  B  distant  x  from  O,  is   W — -, —    , 


FIG.  169. 


and  is  the  Shearing  Force  for  all 
points  between  O  and  B\  it  is  nil  or 
W  according  as  the  weight  is  at  A 
or  0.  Upon  the  vertical  through  O 
take  OD  equal  or  proportional  to  W;  join  DA.  The  shearing 
force  at  any  point  of  the  beam  between  O  and  the  weight,  as 
the  latter  travels  from  A  towards  (9,  is  represented  by  the  ver- 
tical distance  between  that  point  and  the  line  AD. 

Also,  the  shearing  force  at  any  point  between  B  and  A  is 

R^  —  W  =  —  Wjt  and  is  equal  or  proportional  to  the  vertical 

distance  between  that  point  and  the  line  OE  where  AE  is  equal 
to  OD. 

Again,   the   Bending  Moment  at   B,  when   W  is   at  B,  is 

it  is  nil  at  O  and  at  A  ; 

Wl 

it  is  a  maximum  and  = at  the 

4 

middle  point  D.  The  bending  mo- 
ment at  any  point  of  the  beam  when 
the  weight  is  at  that  point  is  repre- 
sented by  the  vertical  distance  be- 
tween the  point  and  the  parabola  OEA,  having  its  axis  vertical 

Wl 

and  its  vertex  at  E,  where  DE  is  equal  or  proportional  to . 

Note. — The  shearing  and  bending  actions  are  symmetrical 
on  both  sides  of  the  centre,  and  it  is  therefore  sufficient  to  deal 
with  one  half  of  the  girder  only. 

Cor.  i.  The  shearing  force  and  bending  moment  at  any 
point  are  maxima  at  the  instant  the  weight  passes  that  point. 

For  example,  the  shearing  force  at  B  for  the  segment  OB, 


FIG.  170. 


112 


THEORY  OF   STRUCTURES. 


when  the  weight  is  at  B,  is  equal  or  proportional  to  BC  (Fig. 
169),  which   is    evidently  greater  than   GH,  representing  the 
shearing  force  at  B,  when  the  weight  is  at  any  other  point  G. 
Again,  the  bending  moment  at  B  (Fig.  170),  when  W  is  at 

B,  is  W — -j — x.     If  W  is  at  any  other  point  G  distant  a  from 


O,   the    bending    moment    at    B    is 


or 


—-t 

according  as  a  <  or  >  x,  and  in  either  case  is  greatest  when 
a  —  x,  i.e.,  when  the  weight  is  at  B. 

Cor.  2.  In  addition  to  the  rolling  load,  let  the  girder  carry 
a  permanent  weight  W  at  the  centre. 

Consider  one  half  of  the  girder  only,  and,  for  convenience, 
trace  the  shearing-force  and  bending-moment  diagrams  for  W 
below  OA. 

The  compound   diagram  for  maximum  shearing  forces  is 

W 
DTLFD  (Fig.  171),  where  KT  is  equal  or  proportional  to  —  , 

W 
and  KL  —  OF  is  equal  or  proportional  to  —  . 

The  maximum  shearing  force  at  a  point  distant  x  from  the 
centre  is  represented  by  XY  '==  -r-(-  -f-  x\  -|  —  —  . 

/    \2  /  2 


F          Y 
FIG.  171. 


Again,  the  compound  diagram  for  maximum  bending  mo- 
ments is  OEFO  (Fig.  172),  where  DF  is  equal  or  proportional 

W'l 
to  ,  and  OF  is  a  straight  line. 


EFFECT  OF  A    ROLLING  LOAD.  113 

The  maximum  bending  moment  at  a  point  distant  x  from 
the  centre  is  represented  by 

XY=m-*\+*L\>— 

/    \4  /  2     \2 

Or.  3.    Theoretically,  the  total  volume  of  material  required 
in  the  web  of  the  girder  in  Cor.  2  is  equal  or  proportional  to 

2  X  ^^DTLF       3  Wl  .    i  W'l 


yj  being  the  web  unit  stress. 

So,  if  </be  the  effective  depth  of  the  girder,  and  f  the  unit 
stress  in  one  of  the  flanges,  the  total  volume  of  metal  in  that 
flange  is  equal  or  proportional  to 

2  X  area  OEFO       2  WF       W'T        I  Wl*        I  W'P 


fd  ~ 

CASE  II.  Let  a  train  weighing  w  per  unit  of  length  travel 
over  the  girder  from  right  to  left,  and  let  the  total  length  o£ 
the  train  be  not  less  than  that  of  p 
the  girder. 

The   reaction    at  A,  Fig.   173, 
when  the  front  of  the  train  is  at 

*7Jt)3u 

B  distant  x  from  O,  is  -^  ,  and 

is  the  shearing  force  for  all  points 

between  A  and  B.     Upon  the  ver-  FlG-  J73. 

ticals  through  A  and  O  take  AD  and  OE  each  equal  or  pro- 

portional to  -  —  .     Thus  between  A  and  B  the  shearing  force 

at  any  point  is  represented  by  the  vertical  distance  between 
that  point  and  a  parabola  having  its  axis  vertical  and  its  vertex 
at  O. 

After  the  end  of  the  train  has  passed  O,  the  shearing  force 
at  any  point  of  the  uncovered  portion  of  the  girder  is  evidently 
represented  by  the  vertical  distance  between  that  point  and  the 
parabola  AFE,  having  its  axis  vertical  and  its  vertex  at  A. 

in,  as  the  train  moves  from  O  towards  B,  the  reaction 


114  f&EORY  OF  STRUCTURES. 

at  A,  and  consequently  the  bending  moment  at  B,  continually 
increase.  On  passing  B,  the  reaction  at  A  still  increases,  and 
the  bending  moment  at  B  when  the  train  covers  a  length  a 
of  the  girder  is 

we?  w  ,  wx 


This  expression  is  evidently  a  maximum  when  a(2l  —a)  is  a 
maximum,  i.e.,  when  a  ^  I.  Hence  the  bending  moment,  and 
therefore  the  flange  stresses,  at  any  point  are  greatest  when  the 
moving  load  covers  the  whole  girder. 

Cor.  I.  The  shearing  force  at  any  point  B  is  a  maximum 
when  the  train  covers  the  longest  segment  OB. 

This  is  evidently  the  case  until  the  train  arrives  at  B,  for 
the  reaction  at  A,  and  therefore  the  shearing  force  at  B,  will 
continually  increase  up  to  this  point.  When  the  train  passes 
B  and  covers  a  length  a(>x)  of  the  girder,  the  shearing  force 

.    wo* 
at  B  is  —  -j  --  w(a  —  x). 

But  this  is  <  —  r  ,  the  shearing  force  at  B  when   OB  is 

.  c?  —  x*  f  a  +  x 

covered,  if   •    —.  —  <  a  —  x,   i.e.,   if  -  -.  —  <  I,   which    is   evi- 

2/  2/ 

dently  the  case. 

Cor.  2.  In  designing  the  flanges  of  a  girder,  the  rolling 
load  is  supposed  to  cover  the  whole  girder,  and  may  be  treated 
as  a  uniformly  distributed  load. 

Cor.  3.  In  addition  to  the  roll- 
ing load,  let  the  girder  carry  a  uni- 
formly distributed  load  of  w'  per 
0          unit  of  length. 


r     G  As  before,  consider  one  half  of 

the  girder  only.     Trace  the  shear- 
ing-force diagram  for  the  perma- 
FlG-  I74'  nent  load  below  OA.     The  com- 

pound diagram  is  DHGK,  where  GH  and  AK  are  equal  or 

wl          w'l 
proportional  to  -~-  and ,  respectively. 


EFFECT  OF  A    ROLLING  LOAD.  115 

The  maximum  shearing  force  at  a  point  distant  x  from  the 
centre  is  represented  by  XY  and  is  equal  to 


wfl_ 

27V  2 


Again,  the  maximum  flange-stresses  are  obtained  by  assum- 
ing the  total  load  upon  the  girder  to  be  w  -f-  w'  per  unit  of 
length. 

Ex.  The  two  main  girders  of  a  single-track  bridge  are  80 
ft.  in  the  clear  and  10  ft.  deep.  The  dead  load  upon  the 
bridge  is  2500  Ibs.  per  lineal  foot.  If  the  bridge  is  traversed  by 
a  uniformly  distributed  live  load  of  3000  Ibs.  per  lineal  foot, 
determine  the  maximum  bending  moment  and  shearing  force 
at  a  point  of  the  girder  distant  10  ft.  from  one  end. 

The  bending  moment  at  any  point  is  a  maximum  when  the 
train  covers  the  whole  of  the  bridge,  in  which  case  the  total 
distributed  load  is  5500  Ibs.  per  lineal  foot,  of  which  each  girder 
carries  one  half. 

Thus  the  reaction  at  each  support  =  — .  80.  ~ —  =  1 10,000 

Ibs.,  and  the  bending  moment  at  the  given  point  =  nooooX  IO 
—  10  X  2750  X  5  =  962,500  Ib.-ft. 

The  shearing  force  at  the  given  point  due  to  the  dead  load 
=  noooo  —  10x2750  —  82,500  Ibs. 

The  shearing  force  due  to  the  live  load  is  a  maximum  when 
the  live  load  covers  the  70  ft.  segment,  and  its  value  is  then 


1 500x70" 

lbs- 


Hence  the  total  maximum  shearing  force 
=  82,500  +  45,937*  =  128,437^ 


THEORY  OF  STRUCTURES. 


5.  Moments  of  Forces  with  respect  to  a  given  Point  Q. 
— First,  consider  a  single  force  /*,. 

Describe   the  force   and   fu- 
.  .... 

nicular   polygons,  i.e.,  the   line 

S,S9  and  the  lines  AB,  BC. 

Through  the  point  Q  draw  a 
line  parallel  to  S,S6,  cutting  the 
lines  AB  and  CB  produced  in  x 
and  y. 

Drop  the  perpendiculars  BM 
and  ON  upon  yx  and  5,56  produced.  Then 


.'.  P,BM  =  xy  .  ON. 

But  .#lf  is  equal  to  the  length  of  the  perpendicular  from  Q 
to  the  line  of  action  of  Plf  and  the  product  xy .  ON  is,  there- 
fore, equal  to  the  moment  of  Pl  with  respect  to  Q.  Hence,  if 
a  scale  is  so  chosen  that  ON  =  unity,  this  moment  becomes 
equal  to  xy ;  i.e.,  it  is  the  intercept  cut  off  by  the  two  sides  of 
the  funicular  polygon  on  a  line 
drawn  through  the  given  point 
parallel  to  the  given  force. 

Next,  let  there  be  two  forces, 
P  P 

*ii  *«• 

Describe  the  force  and  fu- 
nicular polygons  vS^vSe  and 
ABCD. 

Let  the  first  and  last  sides 
(AB  and  DC]  be  produced  to 
meet  in  G,  and  let  a  line  through 
the  given  point  Q  parallel  to  the  line  5aS6  intersect  these  lines 
in  x  and  y. 

Draw  GM  perpendicular  to  xy,  and  ON  perpendicular  to 

•S.S..    Then 

xy     _  S,S,  _  resultant  of  P,  and  Pa 
~GM  =  ON  =  "ON 


MOMENTS   OF  FORCES. 


and  hence 

(the  resultant  of  P,  and  P2)  X  GM =  xy .  ON. 

But  GMis  equal  to  the  length  of  the  perpendicular  from  Q 
upon  the  resultant  of  Pl  and  P^,  which  is  parallel  to  S^S6  and 
must  necessarily  pass  through  G.  Hence,  if  a  scale  is  so  chosen 
that  ON  =  unity,  xy  is  equal  to  the  moment  of  the  forces  with 
respect  to  Q ;  i.e.,  it  is  the  intercept  cut  off  by  the  first  and  last 
sides  of  the  funicular  polygon  on  a  line  drawn  through  the  given 
point  parallel  to  the  resultant  force. 

A  third  force  P3  may  be  compounded  with  Pl  and  P9,  and 
the  proof  may  be  extended  to  three,  four,  or  any  number  of 
forces. 

The  result  is  precisely  trfe  same  if  the  forces  are  parallel. 

The  force  polygon  of  the  n  parallel  forces  Pl ,  P9 ,  .  .  .  PH 


/ 


FIG.  177. 


becomes  the  straight  line  565,52  .  .  .  Sn .  Let  the  first  and  last 
sides  of  the  funicular  polygon  meet  in  G.  Drop  the  perpen- 
diculars GM,  ON  upon  xy  and  S6Sn,  xy,  as  before,  being  the 
intercept  cut  off  on  a  line  through  the  given  point  Q  parallel 
to  SCSH .  Then 

xy.ON  =  GM.  S6Sn .     Hence,  etc. 

Thus  the  moment  of  any  number  of  forces  in  one  and  the 
same  plane  with  respect  to  a  given  point  may  be  represented 
by  the  intercept  cut  off  by  the  first  and  last  sides  of  the  funicu- 


Il8  THEORY  OF  STRUCTURES. 

lar  polygon  on  a  line  drawn  through  the  given  point  parallel  to 
the  resultant  of  the  given  forces. 

6.  Bending  Moments. — Stationary  Loads. — Let  a  hori- 
zontal beam  AB,  supported  at  A  and  B,  carry  a  number  of 
weights  Plt  />,  P».  .  .  at  the  points  N19  N,,  N3,  .  .  . 


I       NX          No     N3     L  N4    N5  M 


FIG.  178. 


The  force  polygon  is  a  vertical  line  1234  .  .  .  n,  where 
12  =/>,  23  =  />,  etc. 

Take  any  pole  (9  and  describe  the  funicular  polygon 
AtA%A^  .  .  . 

Let  the  first  and  last  sides  of  this  polygon  be  produced  to 
meet  in  G  and  to  cut  the  verticals  through  A  and  B  in  the 
points  C  and  D. 

Join  £Z>. 

Let  the  vertical  through  G  cut  AB  in  Z,  and  CD  in  .£"; 
is  the  line  of  action  of  the  resultant. 

Draw  OH  parallel  to  CD. 

From  the  similar  triangles  OiH  and  GCK, 


\H       GK 

~OH  ~  CK  ' 


BENDING  MOMENTS. 

From  the  similar  triangles  OnH  and  GDK, 

nH_       GK 
OH  ~  DK  ' 

iH      DK      BL      R 


R}  ,  /?2  being  the  reactions  at  A  and  B,  respectively. 
But  iH+  nH  =  in  =  P,  +  P,  +  .  .  .  =  R,  +  R,  . 

Hence  iH  =  R,     and     nH  =  R,. 

Thus  the  line  drawn  through  the  pole  parallel  to  the  closing- 
line  CD  divides  the  line  of  loads  into  two  segments,  of  which  the 
one  is  equal  to  the  reaction  at  A  and  the  other  to  that  at  B. 

Let  it  now  be  required  to  find  the  bending  moment  at  any 
point  M  of  the  beam,  i.e.,  the  moment  of  all  the  forces  on  one 
side  of  J/with  respect  to  M. 

In  the  figure  these  forces  are  Rt  ,  Pl  ,  />a  ,  Ps  ,  P4  ,  P6  ,  and  the 
corresponding  force  polygon  is  Hi  23456.  The  first  and  last 
sides  of  the  funicular  polygon  of  the  forces  are  CD  parallel  to 
OH,  and  A^A^  parallel  to  O6.  If  the  vertical  through  M  meet 
these  sides  in  x  and  y,  then,  as  shown  in  Art.  5,  the  moment  of 
the  forces  R^  ,  Pl  ,  P9  ,  Pz  ,  Pt  ,  P6  with  respect  to  M,  i.e.,  the 
bending  moment  at  M,  =  ON  .xy,  ON  being  the  perpendicular 
from  O  upon  iH  produced. 

Hence,  if  a  scale  is  chosen  so  that  the  polar  distance  ON  is 
unity,  the  bending  moment  at  any  point  of  the  beam  is  the  inter- 
cept on  the  vertical  through  that  point  cut  off  by  the  closing  line 
CD  and  the  opposite  bounding  line  of  the  funicular  polygon. 

7.  Moving1  Loads.  —  Beams  are  often  subjected  to  the 
action  of  moving  loads,  as,  e.g.,  in  the  case  of  the  main  girders 
of  a  railway  bridge,  and  it  becomes  a  matter  of  importance  to 
determine  the  bending  moments  for  different  positions  of  the 
loads.  It  may  be  assumed  that  the  loads  are  concentrated  on 
wheels  which  travel  across  the  bridge  at  invariable  distances 
apart. 

At  any  given  moment,  let  the  figure  represent  a  beam  1  1 


120 


THEORY   OF  STRUCTURES. 


under  the  loads  Plt  P9,  P3.  .  .  Describe  the  corresponding 
funicular  polygon  CC'C" .  .  .  D,  the  closing  line  being  CD. 

Let  the  loads  now  travel  from  right  to  left.  The  result  will 
be  precisely  the  same  if  the  loads  remain  stationary  and  if  the 
supports  1 1  are  made  to  travel  from  left  to  right. 

Thus,  if  the  loads  successively  move  through  the  distances 

C' 


FIG.  179. 

12,  23,  34,  ...  to  the  left,  the  result  will  be  the  same  if  the 
loads  are  kept  stationary  and  if  the  supports  are  successively 
moved  to  the  right  into  the  positions  22,  33,  44,  .  .  .  The  new 
funicular  polygons  are  evidently  C' C"  ...£>',  C"C"  ,  .  .  D", 
C'"C""  .  .  .  D'",  ...  the  new  closing  lines  being  CD',  C"D", 
C"D'",  .  .  . 

The  bending  moment  at  any  point  M  is  measured  by  xy  for 
the  first  distribution,  xy'  for  the  second,  x"y"  for  the  third, 
etc.,  the  position  of  M  for  the  successive  distributions  being  de- 
fined by  MM'  =  12,  M'M"  =  23,  M"M'"  =  34,  .  ... 

Similarly,  if  the  loads  move  from  left  to  right,  the  result 
will  be  the  same  if  the  loads  are  kept  stationary  and  if  the  sup- 
ports are  made  to  move  from  right  to  left. 

It  is  evident  that  the  envelope  for  the  closing  line  CD  for 
all  distributions  of  the  loads  is  a  certain  curve,  called  the  enve- 
lope of  moments.  The  intercept  on  the  vertical  through  any 
point  of  the  beam  cut  off  by  this  curve  and  the  opposite  bound- 


MAXIMUM   SHEAR  AND   BENDING  MOMENT. 


121 


ary  of  the  funicular  polygon  is  the  greatest  possible  bending 
moment  at  that  point  to  which  the  girder  can  be  subjected. 

EXAMPLE.  Loads  of  12  and  9  tons  are  concentrated  upon  a 
horizontal  beam  of  12  ft.  span  at  distances  of  3  and  9  ft.  from 
the  right-hand  support.  Find  (a)  the  B.  M.  at  the  middle  point 


12' 


3' 


T 

J9  tor 


!12  tons-'' 


FIG.  180. 

of  the  beam,  and  also  (ft)  the  max,  B.  M.  produced  at  the  same 
point  when  the  loads  travel  over  the  beam  at  the  fixed  dis- 
tances of  6  ft.  apart. 

Scales  for  lengths,  -J  in.  =  I  ft.  ;  for  forces,  -^  in.  —  I  ton. 

Take  polar  distance  =  £  in.  —  10  tons. 

Case  a.  B.M.  =  xy  X  10  =  3.15  X  lOtons  =  31  J  ton-ft. 
Case  b.   B.  ^l.—x'y'  X  10  =  3.6    X  10  tons  =  36  ton-ft. 


8.  Analytical  Method  of  Determining  the  Maximum 
Shear  and  Bending  Moment  at  any  Point  of  an  Arbitrarily 
Loaded  Girder  AB.—  At  any  given  moment  let  the  load  con- 
sist of  a  number  of  weights  «;,,«;„,...  wn,  concentrated  at 
points  distant  al  ,  a^  ,  .  .  .  an  ,  respectively,  from  B. 

The  corresponding  reaction  R^  at  a  is  given  by 


RJ,  —  it>lal 


wa 


/being  the  length  of  the  girder. 

Let  Wn  =  w}  +  w2  +  •  •  •  +  WM,  the  sum  of  the  n  weights. 
"  Wr  =  ^vl  +  w2  +  .  .  .  +  wr  ,  the  sum  of  the  first  r  w'ts. 

The  shear  at  a  point  P  between  the  rth  and  the  (r  -\-  i)th 
weights  is 

5,  =  R,  -  w,  -  w,  —  .  .  .  -  wr  =  R,  —  Wr  . 


122  THEORY  OF  STRUCTURES. 

Let  all  the  weights  now  move  towards  A  through  a  distance 
x,  and  let  p  of  the  weights  move  off  the  girder,  q  of  the  weights 
be  transferred  from  one  side  of  P  to  the  other,  and  s  new 
weights,  viz.,  wn+1  ,  wn+^  ,  .  .  .  wn+s  ,  advance  upon  the  girder, 
their  distances  from  B  being  an+l  ,  an+y  ,  .  .  .  an+s,  respectively. 

Let  L  =  w^  -f-  wz  +  .  .  .  +  wp  ,  the  total  weight  leaving  the 
girder. 

Let  T  =  wr+l  -f-  wr+t  +  .  .  .  +  wr+q  ,  the  total  weight  trans- 
ferred from  one  side  of  P  to  the  other. 

Let  RPl  =  wlal  -f-  wa^2+  •  •  • 
"    Rl  =  wrar          Wr2ar 


q 


Thus  Rp,  Rq,  Rs  are  the  reactions  at  A  due,  respectively,  to 
the  weight  which  leaves  the  girder,  the  weight  which  is  trans- 
ferred, and  the  new  weight  which  advances  upon  the  girder. 

The  reaction  R^  at  A  with  the  new  distribution  of  the  loads 
is  given  by 


RJ  =  wp+l(ap^  +  x)  +  Wp+2(ap+2  +  x)  +  .  .  .  +  wr(ar  +  x) 
+  iVr+^a^  +  x)  +  .  .  .  +  wn(an  +  x)  +  zvn+1an+1  +  ..  . 
+  ow^,  -  Rj  -  RPl  +x(Wn-L)  +  RJ, 

and  hence 


Also,  the  corresponding  j^^r  at  P  is 

wp^  +  .  .  .  +  wr  + 


Hence  the  shear  at  P  with  the  first  distribution  of  weights 
is  greater  or  less  than  the  shear  at  the  same  point  with  the 
second  distribution  according  as 


> 

1   < 


or  ^      rt- 

or  T  -  L  >  R,  - 


MAXIMUM   SHEAR   AND   BENDING  MOMENT.  12$ 

Note. — When  no  weights  leave  or  advance  upon  the  girder, 
Rs ,  RP ,  and  L  are  severally  nil,  and  hence 

.S  >  5" 
according  as  -  ^  — - ; 

i.e.,  according  as  the  weight  transferred  divided  by  the  distance 
through  which  it  is  transferred  is  greater  or  less  than  the  total 
weight  on  the  girder  divided  by  the  span. 

Again,  let  z  be  the  distance  of  P  from  B,  and  let 

T)      7  I  I  I 

J\.rl-  '=•  Wl&l  — |—  W^a^  — f—  .  .  .  — j—  Wf/Zf. . 

The  bending  moment  at  P  with  the  first  distribution   of 
weights  is 

J/j  =  R^(l  —  z)  —  ^(X  —  z)  —  w2(tf3  —  z)  —  ...  —  wr(ar  —  z) 


The  bending  moment  at  the  same  point  with  the  second  dis- 
tribution is 


,  =  R%(1  —z)—  wp+l(a^  +  x  —  z)~  wp+2(ap+2  +  x  —  z)  —  .  .  . 

—  wr(ar  +  x  —  z)  —  .  .  .  —  wr+q(ar+q  +  x  —  z) 
=  RJJL  -z)-  (Rrl  -  RPl  +  Rql)  -(x-z}(Wr-L+  T). 


Hence  the  bending  moment  at  P  with  the  first  distribution 
of  weights  is  greater  or  less  than  the  bending  moment  at  the 
same  point  with  the  second  distribution  according  as 


or 


,(l  -z)-  (Rr  -Xf  +  Rq)l 

-(x-z)(Wr-L+T), 


124  THEORY  Of  STRUCTURES. 

or 

*Wr-  (Rf  -  R,y+  (x  -  *X  W,  -  L  +  T)  >  (R,  -  *,)(/-  *) 

or 


(B) 


.  —  If  no  weights  leave  or  advance  upon  the  girder  Rs, 
and  L  are  severally  nil,  and 


according  as 


If  also  the  point  P  coincide  with  the  rth  weight,  and  the 

distance  of  transfer,  xl  =  ar  —  ar+l  ,  then 

Rql  —  tvr+lar+I  ,     T  —  wr+l  ,     and     z  =  ar  . 
Hence  J/i  ^  J/2  ,  according  as 


or 


i.e.,  according  as  the  sum  of  the  first  r  weights  divided  by  the 
length  of  the  corresponding  segment  is  greater  or  less  than  the 
total  weight  upon  the  girder  divided  by  the  span, 

If  the  weights  are  concentrated  at  the  panel  points  of  a 
truss,  the  last  relation  may  be  expressed  in  the  form 

first  (r)  weights  >  total  weight 

r  panels        <  total  number  of  panels' 

EXAMPLE.  A  series  of  loads  of  3000,  23,600,  20,100,  21,700, 


MAXIMUM    SHEAR   AND   BENDING  MOMENT.  125 

22,900,  18,550,  18,000,  18,000,  and   18,000  Ibs.  travel,  in  order, 
over  a  truss  of  240  ft.  span  and  ten  panels. 

Let  ^AA  .  .  .  B  be  the  truss,  A »  A»  A »  •  •  •  being  the 
panel  points.  Let  the  loads  travel  from  B  towards  A,  and 
compare  the  shear  in  the  panel  AA  when  the  weight  of  3000 
Ibs.  has  reached  A  with  the  shear  in  the  same  panel  when  the 
weights  have  advanced  another  24  ft. 

R'  =  lo  *  I855°  =  l855  lbs"    R>  =  °'    7  =  To 
Wn  —  91300  lbs.,     L  —  o,     T  =  3000  lbs. 
Hence  S^S^  according  as  (see  A) 


3000  -  0^1855  +  -(91300  -  0)>  10985, 


and 


Let  the  weights  again  advance  24  ft. 

Rs  =  --  .  18000  =  1800  Ibs.,     Rp  =  09     X-j  —  --, 
10  /        IO 

Wn  —  109,300  Ibs.,     L  —  o,     T  ~  23,600  Ibs. 
Hence  St  ^J  S2  ,  according  as  (see  A) 
23600  —  o^  1800  —  o  +  -^(109300  —  o),     or     23600^  12730, 

and 

.-.  s,>s,. 

Hence  the  shear  in  the  panel  A  A  'ls  a  maximum  when  the 
weight  of  3000  Ibs.  is  at  A  • 

Again,  let  the  3000  Ibs  be  at  A>  and  compare  the  bending 
moment  at  A  w^h  t^ie  Dencling  moment  at  the  same  point 
when  the  weights  have  advanced  first  24  ft.  and  then  48  ft. 
towards  A. 

First,  z—  120  ft.,  L=o,  T=  22,900  Ibs.,  AV—  18000X24, 


126 


THEORY  OF   STRUCTURES. 


RP  =  o,      Rql  =  22900  X  96>      x  —  24  ft.,     Wr  =  68,400  Ibs., 
Wn  —  145,850  Ibs. 

Hence  M^M^  according  as  (see  B) 


120(0  —  22900  +  JSoo  —  o)  +  22900  X  96  —  18000  X  24 

+  24(68400  -  o  +  22900)  $  -—(240  -  120X145850  -  o), 


240 

or  1425600^1750200, 

and 


Second,  z—  120  ft.,  L  =  3000  Ibs.,  T—  18550)  Rs  =  o,  Rpl 
=  3000  X  216,  Rql=  18550X96,  #  =  24  ft.,  ^  =  91,300  Ibs., 
Wn  =  163,850  Ibs. 

Hence  Mt  ^  J/Q  ,  according  as  (see  B) 

120(3000  —  18550  +  0  —  3000.—  -}  +240(18550.  -^  --  o) 
V  2407  \  240         ' 

+  24(91300—  3000+  18550)^—  -(240-  120X163850  —  3000), 

240 

or  2155200^1930200, 

and 


Hence  the  bending  moment  at  /6  is  a  maximum  when  the 
weight  of  3000  Ibs.  is  at/1?  i.e.,  when  all  the  panel  points  are 
loaded. 

9.  Hinged  Girders.  —  Any  point  of  a  girder  at  which  the 
bending  moment  is  nil  is  termed  a  point  of  contrary  flexure, 
and  on  passing  such  a  point  the  bending  moment  must  neces- 
sarily change  sign. 

Consider  a  horizontal  girder  resting  upon  supports  at  A,  B, 
C,  D,  and  hinged  at  the  points  E  and  F  in  the  side  spans. 

In  order  that  there  may  be  no  distortion  by  the  turning  of 
the  hinges,  the  latter  must  not  be  subject  to  any  bending 
action  ;  i.e.,  they  must  be  points  of  contrary  flexure. 


HINGED    GIRDERS. 


127 


Let  AE  =  a,  EB  =  b,  BC  =  c,  CF  =  e,  DF  =  d. 

Let  W^  W^  Wz,  W^  W,  be  the  loads  upon  AE,  EB,  BC, 
DF,  FC,  respectively,  and  let  xl ,  x^ ,  xz ,  x^ ,  xb  be  the  several 
distances  of  the  corresponding  centres  of  gravity  from  the 
points  E,  B,  C,  F,  C. 

Mt 


FIG.  181. 

The  two  portions  AE  and  DF  are  evidently  in  precisely  the 
same  condition  as  two  independent  girders  of  the  same  lengths, 
carrying  the  same  loads  and  supported  at  the  ends.  EF  may 
also  be  treated  as  an  independent  girder  supported  at  B  and  C, 
carrying  the  weights  J/F2,  Wz,  W&,  and  loaded  at  the  cantilever 
ends  E  and  F  with  weights  equal  to  the  reactions  at  E  and  F 
for  the  portions  AE,  DF  assumed  to  be  independent  girders. 

Let  R^R^R^  R,  be  the  reactions  at  A,  B,  C,  A  respec- 
tively. Then 


and 


Hence,  since  R^  and  R^  are  always  positive,  there  can  be  no 
upward  pull  either  at  A  or  D,  and  no  anchorage  will  be  needed 
at  these  points. 

Next,  taking  EF  as  an  independent  girder, 


the  load  at  E  =  IV,  -  R,  = 


«     F=  W,-R,= 


128  THEORY  OF  STRUCTURES 

Take  moments  about  C  and  B.    Then 

-  (  W,  -  *,X*  +  0  -  Wfo  +  «)  +  ^  -  »>.  + 


and 

-(IV,-  Rtf  -  W^  +  W3(c  -  x)  -  R,c 


two  equations  giving  R^  and  Ra,  since  Rl  and  R4  have  been 
already  determined. 

The  pier  moments  Pl  at  B  and  Pt  at  C  are 


and 
P  = 


, 


'r  values  depending  splely  upon  the  loads  on  the  spans  contain- 
ing the  hinges. 

The  bending  moment  at  any  point  in  BC  distant  x  from  B 

R,)(b  +  x]  -  W&,  +  x)-M 


M  being  the  bending  moment  due  to  the  load  upon  the 
length  x. 

The  shearing-force  and  bending-moment  diagrams  for  the 
whole  girder  can  now  be  easily  drawn. 

For  any  given  loads  upon  the  side  spans,  let  AEH  and  DFL 
be  the  bending-moment  curves  for  the  portions  AB,  CD  ;  BH 
and  CL  representing  the  pier  moments  at  B  and  C,  respec- 
tively. The  bending  moments  for  the  least  and  greatest  loads 
upon  BC  will  be  represented  by  two  curves  HKL,  HK'L,  and 
the  distances  TT't  VV  through  which  the  points  of  contrary 
flexure  must  move,  indicate  those  portions  of  the  girder  which 
are  to  be  designed  to  resist  bending  actions  of  opposite  signs. 


HINGED   GIRDERS. 


I29 


Again,  let  the  two  hinges  be  in  the  intermediate  span. 

Let  AB  =  a,  BE  =  b,  EF  =  c,  FC  =  e,  CD  =  d. 

Let  W,,  W,,  W,,  W,,  W,  be  the  loads  upon  AB,  BE,  EF, 
CD,  CF,  respectively,  and  let  xl ,  #„ ,  xa,  x^,  x&  be  the  several 
distances  of  the  corresponding  centres  of  gravity  from  the 
points  B,  B,  F,  C,  C. 


// 

f 


FIG.  182. 

EF  evidently  may  be  treated  as  an  independent  girder  sup- 
ported at  the  two  ends  and  carrying  a  load  W3. 

AE  and  DF  may  be  treated  as  independent  girders  carry- 
ing the  loads  Wl ,  W^  and  W^ ,  W^,  respectively,  and  also  loaded 
at  the  cantilever  ends  E  and  F  with  weights  equal  to  the  reac- 
tions at  E  and  F  due  to  the  load  Wa  upon  girder  EF,  which  is 
assumed  to  be  independent.  Thus 


the  load  at  E  =  W,-  ; 


The  pier  moments  Pl  at  B  and  P9  at  C  are 


and 


their  values  depending  solely  upon  the  loads  on  the  span  contain- 
ing the  hinges. 


13°  THEORY  OF  STRUCTURES. 

Let  R,,  R,,  Ra,  Rt  be  the  reactions  at  A,  B,  C,  D,  respec 
tively,  and  take  moments  about  the  points  B,  A,  D,  C.    Then 


R     - 


R,d~  W,i  -       (,  +  d)-  W.(xt  +  d)  -  Wt(d  -  xt)  =  O 
-  Rtd  -  W,i  -  $e  -  WtXi  +  Wtx.  =  o 


Rn  and  ^8  are  always  positive; 

R^  is  positive  or  negative  according  as  W^x,  ^  Pl  ;  and 

t>  t(  tt  tt  "         W  r   >  P 

A«  VV  1X*<L1    2  • 

Thus  there  will  be  a  downward  pressure  or  an  upward  pull 
at  each  end  according  as  the  moment  of  the  load  upon  the  ad- 
joining span  is  greater  or  less  than  the  corresponding  pier  mo- 
ment. The  ends  must  therefore  be  anchored  down  or  they 
will  rise  off  their  supports. 

The  shearing-force  and  bending-moment  diagrams  for  the 
whole  girder  can  now  be  easily  drawn. 

Let  HEFL  be  the  bending  moment  curve  for  any  given 
load  upon  the  span  BC,  BH  and  CL  being  the  pier  moments 
at  B  and  C,  respectively. 

The  bending-moment  curves  for  the  least  and  greatest 
loads  on  the  side  spans  may  be  represented  by  curves  A  777, 
*  A  T'H  and  DVL,  DV  L,  and  the  distances  TT,  W  through 
which  the  points  of  contrary  flexure  move  indicate  those  por- 
tions of  the  girder  which  are  to  be  designed  to  resist  bending 
actions  of  opposite  signs. 

Reverse  strains  may,  however,  be  entirely  avoided  by 
making  the  length  of  EF  sufficiently  great  as  compared  with 
the  lengths  of  the  side  spans. 

The  preceding  examples  serve  to  illustrate  the  mechanical 
principles  governing  the  stresses  in  cantilever  bridges. 


EXAMPLES.  13 l 


EXAMPLES. 

1.  A  beam  20  ft.  long  and  weighing  20  Ibs.  per  lineal  foot  is  placed 
upon  a  support  dividing  it  into  segments  of  16  and  4  ft.,  and  is  kept 
horizontal  by  a  downward  force  P  at  the  middle  point  of  the  smaller 
segment.     Find  the  value  of  P  and  the  reaction  at  the  support. 

Show  that  the  required  force  P  will  be  doubled  if  a  single  weight  of 
150  Ibs.  is  suspended  from  the  end  of  the  longer  segment.  Draw  shear- 
ing-force and  bending  moment  diagrams  in  both  cases. 

Ans.   1200  Ibs. ;  1600  Ibs. 

2.  A  man  and  eight  boys  carry  a  stick  of  timber,  the  man  at  the  end 
and  the  eight  boys  at  a  common  point.     Find  the  position  of  this  point, 
if  the  man  is  to  carry  twice  as  much  as  each  boy. 

Ans.  Distance  between  supports  =  f  length  of  beam. 

3.  A  timber  beam  is  supported  at  the  end  and  at  one  other  point ;  the 
reaction  at  the  latter  is  double  that  at  the  end.     Find  its  position. 

Ans.  Distance  between  supports  =  f  length  of  beam. 

4.  Two  beams  ABC,  BCD  are  bolted  at  B  and  C  so  as  to  act  as  one 
beam  supported  at  A  and  D\  AB  —  12  ft.,  £C  =  4  ft.,  CD  =  16  ft, ;  each 
of  the  bolts  will  bear  a  bending  moment  of  100  Ib.-ft.      Find  the  greatest 
weight  which  can  be  concentrated  on  the  portion  BC.     Draw  diagrams 
of  maximum  shearing  force  and  bending  moment  when  a  wheel  of  the 
same  weight  rolls  over  the  beam. 

Ans.  I4T7¥  Ibs. 

5.  In  the  preceding  question  find  the  greatest  uniformly  distributed 
load  which  the  beam  will  bear. 

Draw  the  shearing-force  and  bending-moment  diagrams. 

Ans.  25§f  Ibs. 

6.  A  uniform  beam  20  1/3  ft.  in   length  rests  with  one  end  on  the 
ground  and  the  other  against  a  smooth  vertical  wall ;  the  beam  is  inclined 
at  60°  to  the  vertical  and  has  a  joint  in  the  middle  which  can  bear  a 
bending  moment  of  30,000  Ib.-ft.     Find  the  greatest  load  which  may  be 
uniformly  distributed  over  the  beam.     Also  find  how  far  the  foot  of  the 
beam  should  be  moved  towards  the  wall  in  order  that  an  additional 
2000  Ibs.  may  be  concentrated  at  the  joint. 

Draw  curves  of  shearing  force  and  bending  moment  in  each  case. 

Ans.  8000  Ibs. ;  distance  =  10  ft. 


I32  THEORY  OF  STRUCTURES. 

7.  A  man  of  weight  W  ascends  a  ladder  of  length  /  which  rests 
against  a  smooth  wall  and  the  ground  and  is  inclined  to  the  vertical  at 
an  angle  a.  The  ladder  has  n  rounds.  Find  the  bending  moment  at 
the  rth  round  from  the  foot  when  the  man  is  on  the^th  round  from  the 
foot.  (Neglect  weight  of  ladder.) 

AnSm      M/  *  sin  "' 


8.  A  regular  prism  of  weight  W  and  length  a  is  laid  upon  a  beam  of 
length  2/(>tf).  If  the  prism  is  so  stiff  as  to  bear  at  its  ends  only,  show 
that  the  bending  action  on  the  beam  is  less  than  if  the  bearing  were  con- 
tinuous from  end  to  end  of  the  prism. 

Ans.— ist.  Max.  B.M.  =  W(-  —  - 


2d. 


=  w(l--a\ 

\2        &) 


9.  A  railway  girder,  50  ft.  in  the  clear  and  6  ft.  deep,  carries  a  uni- 
formly distributed  load  of  50  tons.     Find  the  maximum  shearing  stress 
at  20  ft.  from  one  end,  when  a  train  weighing  \\  tons  per  lineal  foot  crosses 
the  girder. 

Also  find  the  minimum  theoretic  thickness  of  the  web  at  a  support 
4  tons  being  the  safe  shearing  inch-stress  of  the  metal. 

Ans.   i6Jtons;  .195  in. 

10.  A  beam  is  supported  at  one  end  and  at  a  second  point  dividing  its 
length  into  the  segments  /«  and  n.     Find  the  two  reactions.     Also  find 
the  ratio  of  »*  to  n  which  will  make  the  maximum  positive  moment  equal 
to  the  maximum  negative  moment. 

Ans,  —  (w2  -  «2),  —  (ni  +  ri)*  ;  m  :  n  :  :  i  +  4/3  :  \/T>. 

27/2  2in 

11.  One  of  the  supports  of  a  horizontal  uniformly  loaded  beam  is  at 
the  end.     Find  the  position  of  the  other  support  so  that  the  straining  of 
the  beam  may  be  a  minimum. 

length 

Ans.  Distance  from  end  support  =  —  =b-. 

•     4/2 

12  A  rolled  joist  17  ft.  long  is  supported  at  one  end  and  at  a  point 
13  ft.  distant  from  that  end.  Two  wagon-wheels  5  ft.  apart  and  each 
carrying  a  load  of  1300  Ibs.  pass  over  the  joist.  Find  the  maximum 
positive  and  negative  moments  due  to  these  weights,  and  also  the  corre- 
sponding reactions. 

Ans.  Max.  positive  B.  M.  =  5512^  Ib.-ft.  ; 

reactions  =  1550  and  1050  Ibs. 
Max.  negative  B.  M.  =  5200  Ib.-ft.  ; 

reactions  =  1700  Ibs.  and  —  400  Ibs. 
or  =  2900  Ibs.  and  —  300  Ibs. 


EXAMPLES.  133 

Denoting  the  distance  from  a  support  by  x,  the  max.  positive  B.  M. 
diagram  for  each  half  of  the  13-ft.  span  is  given  by  Mx  =  100(21  —  2.x)x. 

13.  A  uniformly  loaded  beam  rests  upon  two  supports.     Place  the 
supports  so  that  the  straining  of  the  beam  may  be  a  minimum. 

Ans.  Distance  of  each  support  from  centre  =  /(  i — - 1. 

14.  Two  bars  AC,  CB  in  the  same  horizontal  line  are  jointed  at  C  and 
supported  upon  two  props,  the  one  at  A,  the  other  at  some  point  in  CB 
distant  x  from  C.     The  joint  C  will  safely  bear  n  Ib.-ft.  ;  the  bars  are 
each  /ft.  in  length  and  w  Ibs.  in  weight.     Find  the  limits  within  which 
x  must  lie. 

ivl  ±  2« 

Ans.  I —       — . 
yvl  T  2« 

15.  A  uniform  load  PQ  moves  along  a  horizontal  beam  resting  upon 
supports  at  its  ends  A  and  B.     Prove  that  the  bending  moment  at  a 
given  point  O  is  a  maximum  when  PQ  occupies  such  a  position  that 
OP  :  OQ  : :  OA  :  OB. 

Draw  curves  of  maximum  shearing  force  and  bending  moment  for  all 
points  of  the  beam. 

16.  A  beam  is  supported  at  the  ends  and  loaded  with  two  weights 
m  W  and  n  W  at  points  distant  a,  b,  respectively,  from  the  consecutive 
supports.     Show  that   the   bending  action    is   greatest  at  m  W  or  n  W 

m  >  b 

according  as  -  ^  — . 
n  "^  a 

17.  A  wheel  supporting  10  tons  rolls  over  a  beam  of  20  ft.  span.    Place 
the  wheel  in  such  a  position  as  to  give  the  maximum  bending  moment, 
and  find  its  value. 

Ans.  At  the  centre  ;  50  ton-ft. 

1 8.  Two  wheels  a  ft.  apart  support,  the  one  m  W  tons,  the  other 
n  futons,  m  being  >  n,  and  roll  over  a  beam  of  /  ft.  span.     Show  that 
the  bending  moment  is  an  absolute  maximum  at  the  centre  or  at  a  point 

whose  distance  from  the  nearest  support  is according  as 

2       2(m  +  n) 


^  a\  i  +  |/ ),  and  find  its  value  in  each 

•>     \          '    m  +  nj 


case. 
mWl 


m  Wl          .       m  +  n  rir  (  na      )  2 

Ans. ton-ft.  ; --  W  \  I I     ton-ft. 

4  4/  (         m  +  n  j 

19.  Find  the  max.  B.  M.  on  a  horizontal  beam  of  length  /  supported 
at  the  two  ends  and  carrying  a  load  which  varies  in  intensity  from  w  at 
one  end  to  w  +  px  at  the  other. 


134  THEORY  OF  STRUCTURES. 

20.  Four  wheels  each  carrying  5  tons  travel  over  a  girder  of  24  ft. 
clear  span  at  equal  distances  4  ft.  apart.     Determine,  graphically,  the 
max.  B.  M.  at  8  ft.  from  a  support,  and  also  the  absolute  max.  B.  M.  on 
the  girder. 

Ans.  5p  ton-ft.  ;  80  ton-ft. 

21.  Two  wheels  each  supporting   7  tons  roll   over  a  beam  of  7$  ft. 
span.     Find  the  maximum  bending  moment  for  the  whole  span,  and  also 
the  curve  of  the  maximum  bending  moment  at  each  point  when  the 
wheels  are  4  ft.  apart. 

Ans.  Abs.  max.  B.  M.  =  -^V-  ton-ft.  at  wheel  at  2f  ft.  from  one 
end.  Denoting  the  distance  from  support  by  x,  the  max. 
B.  M.  curve  for  the  first  3^  ft.  is  given  by 

Mx  =  1fi(ll  -2X)X, 

and  for  the  remaining  4  ft.  by 


22.  Two  wheels  supporting,  the  one  1  1  tons,  the  other  7  tons,  travel 
over  a  beam  of  12^  ft.  span.  Find  the  maximum  bending  moment  for  the 
whole  span,  and  also  the  curves  of  the  max.  shearing  force  (both  positive 
and  negative)  and  maximum  bending  moment  at  each  point  when  the 
wheels  are  6  ft.  apart. 

Ans.  Abs.  max.  B.  M.  =  37.2  ton-ft. 

The  max.  positive  shearing  force  at  each  point  is  given  by 
the  equations 

183-18^-  7(121-5) 

s*=—       and  s*=  -—  • 


The  max.  negative  shearing  force  at  each  point  is  given  by 
the  equations 

ix  45i-  1  8*       ~  42  +  i%x  iix 

"'  "'  ~' 


The  max.  B.  M.  curve  is  given  by  the  equations 

183-18* 
Mx  =     ---*     and     Mx= 


N.B.  —  In  the  above  cases  x  is  measured  from  the  support  to  the 
nearest  load. 

23.  In  the  preceding  question  show  that  the  maximum  negative  shear 
at  4^  ft.  from  a  support,  when  the  7-ton  wheel  only  is  on  the  beam,  is  the 
same  as  the  maximum  negative  shear  at  the  same  point  when  both  of 


EXAMPLES.  135 

the  wheels  are  on  the  beam,  and  find  its  value.  Also  show  that  the 
maximum  negative  shear  at  9f  ft.  from  a  support  is  the  same  when  only 
the  1 1- ton  wheel  is  on  the  beam  as  when  the  two  wheels  are  on  the 
beam,  and  find  its  value. 

Ans.  ff§-  tons  ;  ±ffl-  tons. 

24.  Solve  question  22  when  the  beam  carries  an  additional  load  of 
1250  Ibs.  (=  %  ton)  per  lineal  foot. 

Ans.  Abs.  max.  B.  M.  is  at  5.284  ft.  f  = ft.)  from  support. 

Max.  positive  shearing- force  diagram  is  given  by  Sx  = 
18.54625  — 2.065.*- from  x=o\,ox=6%it.,  and  6^=14.90625  — 
1.505-*-  from  x  =  6$  to  x—  12^  ft.  The  max.  negative 
shearing-force  diagram  is  given  by  Sx  =  —  .56^:  from  x  =  o 
to  x  =  4/^  ft-;  =  3.64  —  1.44^1:  from  x  =  4^-  to  x  =  6%  ft.  ; 
=  7.54625  —  2.065.*-  from  x  —  6i  to  x  =  6%  ft. ;  =  3.90625  — 
1.505.*  from  x  =  6%\.ox  =  9f  ft.;  =  9.18625  —  2.065.*:  from 
x  =  9f  to  x  =  I2£  ft.  Max.  B.  M.  curve  is  given  by  Mx  — 
(18.54625  —  1.7525.*-)^,  and  Mx  =  (14.90625  —  1.1925.*-).*-. 

25.  Three  wheels,  each  loaded  with  a  weight  Wand  spaced  5  ft.  apart, 
roll  over  a  beam  of  12  ft.  span.     Place  the  wheels  in  such  a  position  as  to 
give  the  maximum  bending  moment,  and  find  its  value. 

Ans.   Middle  weight  at  centre  of  beam  ;  4  W. 

26.  Place  (a)  the  wheels  in  the  preceding  question  so  that  B.M.  at 
any  point  between  the  two  hindmost  wheels  may  be  constant,  and  find 
its  value.     Also  (8)  determine  all  the  positions  of  the  wheels  which  will 
give  the  same  bending  moment  at  6  and  12  ft.  from  one  end,  and  find  its 
value. 

das. — (a)  ist  wheel  at  i  ft.  from  support;  B.  M.  =  7  W. 

(b)  When   distance    between   end  wheel   and    sup- 
port is  ^  2  ft.  and  ^  5  ft.;  B.  M.  =  7  W. 

27.  Four  wheels  each  loaded  with  a  weight  Wand  spaced  5  ft.  apart 
roll  over  a  beam  of  i8ft.  span.    Place  the  wheels  in  such  a  position  as  to 
give  the  maximum  bending  moment,  and  find  its  value. 

Ans.  One  wheel  off  the  beam  and  middle  wheel  of  remaining 
three  at  the  centre  ;  max.  B.  M.  =  8£  W.  If  all  wheels 
are  on  beam,  max.  B.  M.  =  8  W. 

28.  All  the  wheels  in  the  preceding  question  being  on  the  beam,  the 
B.  M.  at  the  centre  for  a  certain  range  of  travel  is  constant  and  equal  to 
that  for  a  particular  distribution  of  the  wheels  when  only  three  are  on 
the  beam.      Find  the  range,  the  B.  M.,  and  the  position   of  the  three 
wheels. 


136  THEORY  OF  STRUCTURES. 

Ans.  While  the  end  wheel  travels  3  ft.  from  the  support  ;  8  W\ 
first  wheel  5  ft.  from  the  support. 

29.  A  span  of  /  ft.  is  crossed  by  two  cantilevers  fixed  at  the  ends  and 
hinged  at  the  centre.  Draw  diagrams  of  shearing  force  and  bending 
moment  (i)  for  a  single  weight  W  at  the  hinge,  (2)  for  a  uniformly  dis- 
tributed load  of  intensity  w. 

Ans.  Taking  hinge  for  origin,  the  shearing-force  and  bending- 
moment  diagrams  are  given  by 


12) 


30.  A  beam  for  a  span  of  100  ft.  is  fixed  at  the  ends.     Hinges  are  in- 
troduced at  points  30  ft.  from  each  end.     Draw  curves  of  shearing  force 
and   bending  moment  (i)  when  a  weight  of  5  tons  is  concentrated  on 
each  hinge  ;  (2)  when  a  uniformly  distributed  load  of  -J-  ton  per  lineal  foot 
covers  (a)  the  centre  length,  (fr)  the  two  side  lengths,  (c}  the  whole  span. 

Ans.  Take  a  hinge  as  origin  ;  the  diagrams  are  given  by — 
(i)     For  each  side  span        Sx  =  5,  Mx  =  —  $x\ 
for  centre  span       Sx  —  o,  Mx  =  o. 

(2) — (a)  For  each  side  span  Sx  =  —  ,  Mx  = x  ; 

•'•••'•                       .  •                                 5       x                 5          x* 
for  centre  span     Sx  =  —  —  ~  ,   Mx  =  -x >  . 

X  1C  X*^ 

(b)  For  each  side  span  Sx  —  ^-,  Mx  =  —  — x  +  -,  ; 

o  A.  I O 

for  centre  span     Sx  =  o,  Mx  —  o. 

C          Jf  2  C  3r^ 

(c}  For  each  side  span  Sx  =  —  +  5- ,  Mx  — x  +  —  ; 

2          o  4  IO 

5       x  5         jr2 

for  centre  span      Sx  =  —  —  ^- ,   Mx  =  —  x 7  , 

28  2          16 

31.  If  the  load  on  each  of  the  wheels  in  question  27  is  5  tons,  and  if 
the  beam  also  carries  a  uniformly  distributed  load  of  20  tons,  and  two 
loads  of  2  and  3  tons  concentrated  at  points  distant  5  and  9  ft.,  respec- 
tively, from  one  end,  find  the  maximum  shearing  force  (both  positive 
and  negative)  and  the  maximum  bending  moment  for  the  whole  span  ; 
also  find  the  loci  for  the  maximum  shearing  force  and  bending  moment 
at  each  point. 


EXAMPLES.  137 

Ans.  Denoting  the  distance  from  support  by  x,  the  max. 
positive  shearing-force  diagram  is  given  by  Sx  = 
$/-  —  ±g-x  from  x  =  o  to  x  =  3 ;  Sx  =  -*-/-  —  %x 
from  x  =  3  to  x  =  8  ;  S*  =  ±&-  —  %x  from  x  =  8  to 

;r  =  13  ;  S*  =  5  —  ~  from  x  =  13  to  x  =  18  ft.     The 

max.  negative  shear  ing- force  diagram  is  given  by 
Sx  —  —  TVr  from  x  =  o  to  x  =  5  ;  S*  —  f f  —  f ^r  from 
;r  =  5  to  x  —  10 ;  Sr  =  V  —  I-*"  from  x  =  10  to  .r  =  1 5  ; 

50        2O;r 
5^  =  -T-  —  — TT  from  .r  =  15  to-r=  18.     Max.  positive 

shear  =  -3/-  tons  ;  max.  negative  shear  =  $/-  tons  ;  max. 
bending -moment  curve  is  given  by  Mx  =  ^/-x  —  ^-x* 
from  x  =  o  to  x  —  3  ;  Mx  =  ^%±x  —  f  f x"1  from  x  =  3  to 
x  =  5  ;  Mx  =  %-^x  —  Zf-x*  —  1 5  from  x  —  5  to  x  =  8  ; 
^T/r  =  sfy-.r  —  4| jf2  +  12  from  x  =  8  to  x  —  9 ;  abs.  max. 
B.  M.  =  142  ton. -ft. 

32.  A  rolled  joist  weighing  150  Ibs.  per  lineal  foot    and  20  ft.  long 
carries  a  uniformly  distributed  load  of  6000  Ibs.,  and  two  wheels  5  ft. 
apart,  the  one  bearing  5000  Ibs.  and  the  other  3000  Ibs.,  roll  over  the 
joist.     Find  the  maximum  shears  at  the  supports,  at  the  centre,  and  at  5 
ft.  from  each  end. 

Ans.  10,250  Ibs. ;  9750  Ibs. ;  3250  Ibs. ;  6750  Ibs.;  6250  Ibs. 

33.  A  beam  /ft.  long  and  weighing  w  Ibs.  per  lineal  foot  has  a  load 
of  m  W  Ibs.  at  a  ft.  from  the  left  end  and  a  load  of  n  W  Ibs.  at  b  ft. 
from  the  right  end.     Find  the  shearing  forces  and  bending  moments  at 

the  weights  and  at  the  middle  of  the  beam,  a  and  b  being  each  <  — . 
How  will  the  result  be  affected  if  b  >  —  ? 

34.  A  rolled  joist  weighing  450  Ibs.  per  lineal  foot  and  20   ft.  long 
carries  the  four  wheels  of  a  locomotive  at  3,  8,  13,  and   18  ft.  from  one 
end.     Find  the  maximum  bending  moment  and  the  maximum  shears, 
both  positive  and  negative,  the  load  on  each  wheel  being  10,000  Ibs. 

Ans.  Max.    B.  M.  =  102,000  Ib.-ft.  ;   max.    shears  —  19,000  Ibs. 
and  21,000  Ibs. 

35.  Solve  the  preceding  question  when  a  live  load  of  2|  tons  per 
lineal    foot    is   substituted  for  the  four  concentrated   weights  on  the 
wheels. 

36.  The  loads  on  the  wheels  of  a  locomotive  and  tender  passing  over 
a   beam    of  60  ft.  span    are    14,180,  14,180,  21,260,  21,260,  21,260,  21,260, 
16,900,  16,900,  16,900,  16,900  Ibs.,  counting  in  order  from  the  front,  the 


IS  THEORY   OF  STRUCTURES. 

intervals   being  5,  5!,  5,  5,  5,  Sf,  5,  4,  5  ft.      Place  the  wheels  in  such  a 
position  as  to  give  the  maximum  bending  moment,  and  find  its  value. 

Also  find  the  maximum  bending  moments  for  spans  of  30,  20,  and  16 
feet. 

Ans.  For  6o-ft.  span,  max.  B.  M.  is  at  5th  wheel  and  =  1,559,925.4 
Ib.-ft.  when  ist  wheel  is  7.95  ft.  from 
support. 

For  3o-ft.  span,  max.  B.  M.  at  5th  wheel  when  2d  wheel  is 
.596  ft.  from  support  and  =  436,761.4 
Ib.-ft. 

For  2o-ft.  span,  max.  B.  M.  at  centre  when  3d  wheel  is  2£ 
ft.  from  support  and  =  212,600  Ib.-ft.  = 
max.  B.M.  at  same  point  when  4th  wheel 
is  5  ft.  from  support. 

For  i6-ft.  span,  max.  B.  M.  is  at  5th  wheel  and  =  132,875 
Ib.-ft.  when  4th  wheel  is  5  ft.  from  sup- 
port. 

37.  If  the  6o-ft.  beam  in  the  preceding  question  also  carries  a  uni- 
formly distributed  load  of  60,000    Ibs.,   find   the  curves   of    maximum 
shearing  force  and  bending  moment  at  each  point. 

38.  If  a  beam  is  supported  at  the  ends  and  arbitrarily  loaded,  show 
that  the  ordinate  at  the  point  of  maximum  moment  divides  the  area  of 
the  curve  of  loads  into  two  parts  which  are  equal  to  the  supporting 
forces.     If  a  and  b  are  the  distances  of  the  centres  of  gravity  of  the  parts 
from  the  ends  of  the  beam,  and  if  W  is  the  total  weight  on  the  beam, 


show  that  the  maximum  bending  moment  is  W  -s-  I  —  -f  -r- 

39.  A  span  of  /  ft.  is  crossed  by  a  beam   in   two  half-lengths,  sup- 
ported at  the  centre  by  a  pier  whose  width  may  be  neglected.     The  suc- 
cessive weightson  the  wheels  of  a  locomotive  and  tender  passing  overthe 
beam  are  14,000,  22,000,  22,000,  22,000,  22,000,  14,000,  14,000,  14,000,  14,000 
Ibs.,  the  intervals  being  7^,  4$,  4^,  4i,  loj,  5,  5,  5  ft.     Place  the  wheels  in 
such  a  position  as  to  throw  the  greatest  possible  weight  upon  the  centre 
pier,  and  find  the  magnitude  of  this  weight  for  spans  of  (i)  50  ft.;  (2)  25 
ft.;  (3)  20  ft.;  (4)  18  ft. 

40.  Loads  of  3!,  6,  6,  6,  and  6  tons  follow  each  other  in  order  over  a 
ten-panel  truss  at  distances   of  8,  5f,  4^,  and   4|  ft.  apart.     Apply  the 
results  of  Art.  8  to  determine  the  position  of  the  loads  which  will  give 
the  maximum  diagonal  and  flange  stresses  in  the  third  and  fourth  panels. 

41.  A  truss  of  240  ft.  span  and  ten  panels,  has  loads  of  12^,  10,  12,  u, 
9,  9,  9,  9,  and  9  tons  concentrated  at  the  panel   points.     Find,  by  scale 
measurement,  the  bending  moments  at  the  four  panel  points  which    are 
the  most  heavily  loaded,  and  determine  by  Art.  8  whether  these  are  the 


EXAMPLES.  139 

greatest  bending  moments  to  which  the  truss  is  subjected  as  the  weights 
travel  over  the  truss  at  the  panel  distances  apart. 

42.  Loads  of  7i,  12,12,12,12.  tons  are  concentrated  upon  a  horizontal 
beam   of   25   ft.    span   at  distances   of    18,    108,  164,  216,  and    272    in., 
respectively,    from    the   left   support.       Find,  graphically,  the   bending 
moment  at  the  centre  of  the  span.     If  the  loads  travel  over  the  truss  at 
the  given  distances  apart,  find  the  maximum  B.  M.  at  the  same  section. 

43.  A  beam  ABCD  is  supported  at  four  points  A,  B,  C,  and  D,  and 
the  intermediate  span  BC  is  hinged  at  the  two  points  E  and  F.     The 
load  upon  the  beam  consists  of  15  tons  uniformly  distributed  over  AB, 
10  tons  uniformly  distributed  over  BE,  5  tons  uniformly  distributed  over 
FC,  30  tons  uniformly  distributed  over  CD,  and  a  single  weight  of  5  tons 
at  the  middle   point  of  EF.     AB  =  15  ft.;   BE  =5  ft.;  EF  =  15  ft., 
FC  =  10  ft. ;  CD  =  25  ft.     Draw  curves  of   B.  M.  and  S.  F.,  and  find 
points  of  inflexion. 

44.  Four  wheels  loaded  with  4,  4,  8,  and  8  tons  are  placed  upon  a 
girder  of  24  ft.  span  at  distances  of  3  in.,  6£  ft.,  8f  ft.,  and  9  ft.  from  the 
left  support.     Find  by  scale  measurement  the  bending  moment  at  the 
centre  of  the  girder.     If  the  wheels  travel  over  the  girder  at  the  given 
distances  apart,  find  the  maximum  B.  M.  to  which  the   girder  is  sub- 
jected. 

45.  Three  wheels  loaded  with  8.  9,  and  10  tons  and  spaced  5  ft.  apart, 
are  placed  upon  a  beam  of  15  ft.  span,  the  8-ton  wheel  being  3  ft.  from 
the  left  abutment.     Determine  graphically  the  B.  M.  at  6  ft.  from  the 
left  abutment.     Also  find  the  greatest  B.  M.  at  the  same  point  when  the 
weights  travel  over  the  beam,  and  the  abs.  max.  bending   moment   to 
which  the  beam  is  subjected. 

Am.  47f  ton-ft.;  53^  ton-ft.;  abs.  max.  B.  M.  =  56||^  ton-ft.  at 
2d  wheel  when  ist  is  2-§-£-  ft.  from  support. 


CHAPTER   III. 
DEFINITIONS  AND  GENERAL  PRINCIPLES. 

I.  Definitions. — The  science  relating  to  the  strength  of 
materials  is  partly  theoretical,  partly  practical.  Its  primary 
object  is  to  investigate  the  forces  developed  within  a  body,  and 
to  determine  the  most  economical  dimensions  and  form,  con- 
sistent with  stability,  of  that  body.  Certain  hypotheses  have 
to  be  made,  but  they  are  of  such  a  nature  as  always  to  be  in 
accord  with  the  results  of  direct  observation. 

The  materials' in  ordinary  use  for  structural  purposes  may 
be  termed,  generally,  solid  bodies,  i.e.,  bodies  which  offer  an  ap- 
preciable resistance  to  a  change  of  form. 

A  body  acted  upon  by  external  forces  is  said  to  be  strained 
or  deformed,  and  the  straining  or  deformation  induces  stress 
amongst  the  particles  of  the  body. 

The  state  of  strain  is  simple  when  the  stress  acts  in  one 
direction  only,  and  the  strain  itself  is  measured  by  the  ratio  of 
the  deformation  to  the  original  length. 

The  state  of  strain  is  compound  when  two  (or  more)  stresses 
act  simultaneously  in  different  directions. 

A  strained  body  tends  to  assume  its  natural  state  when  the 
straining  forces  are  removed :  this  tendency  is  called  its  elas- 
ticity. A  thorough  knowledge  of  the  laws  of  elasticity,  i.e.,  of 
the  laws  which  connect  the  external  forces  with  the  internal 
stresses,  is  absolutely  necessary  for  the  proper  comprehension 
of  the  strength  of  materials.  This  property  of  elasticity  is  not 
possessed  to  the  same  degree  by  all  bodies.  It  may  be  almost 
absolute,  or  almost  zero,  but  in  the  majority  of  cases  it  has  a 
mean  value.  Hence  it  naturally  follows  that  solid  bodies  may 
be  classified  between  two  extreme,  though  ideal,  states,  viz., 

140 


ELEMENTARY  PRINCIPLES   OF  ELASTICITY.  14! 

a  perfectly  elastic  state  and  a  perfectly  soft  state.  Perfectly 
elastic  bodies  which  have  been  strained  resume  their  original 
forms  exactly  when  the  straining  forces  are  removed.  Per- 
fectly soft  bodies  are  wholly  devoid  of  elasticity  and  offer  no 
resistance  to  a  change  of  form. 

Bodies  capable  of  undergoing  an  indefinitely  large  deforma- 
tion under  stress  are  said  to  be  plastic. 

2.  Stresses    and    Strains.  —  Every    body    may    be    sub- 
jected to  five  distinct  kinds  of  stresses,  viz.  : 

(a)  A  longitudinal  pull,  or  tension. 

(b)  A  longitudinal  thrust,  or  compression. 

(c)  A  shear,  or  tangential  stress,  which  may  be  defined  as  a 
stress   tending  to  make   one  surface  slide   over  another  with 
which  it  is  in  contact. 

(d)  A  transverse  stress. 

(e)  A  twist  or  torsion. 

Under  any  one  of  these  stresses  a  body  may  suffer  either  an 
elastic  deformation,  of  a  temporary  character,  or  a  plastic  de- 
formation, of  a  permanent  character. 

3.  Resistance  of  Bars  to  Tension  and  Compression.  — 
Let  a  straight  bar  of  homogeneous  material  and  length  L  be 
stretched  or  compressed  longitudinally  by  a  force  P  uniformly 
distributed  over  the  constant  cross-section  A   of  the  bar  ;  let 
the  line  of  action  of  P  coincide  with  the  axis  of  the  bar,  and 
let  /  be  the  consequent  extension  or  compression,  i.e.,  the  de- 
formation. 

If  the  transverse  dimensions  of  the  bar  are  small  as  com- 
pared with  the  length,  experiment  shows  that,  within  certain 
limits,  the  force  Pis  directly  proportional  to  the  deformation  / 
and  to  the  area  A,  and  inversely  proportional  to  the  length  L, 
these  quantities  being  connected  by  the  relation 


, 

where  E  is  a  constant  dependent  upon  the  material  of  the  bar 
and  is  called  the  coefficient  or  modulus  of  elasticity.  It  is  evi- 
dently the  force  which  will  double  the  length  of  a  perfectly 

i  p\ 
elastic  bar  of  unit  section.     Denoting  the  unit  stress  L—rJ  by/, 

\Ar 


142  THEORY  OF  STRUCTURES. 

and  the  strain  per  unit  of  length  (—  J  by  A,,  the  above  equation 
may  be  written 


or  the  unit  stress  =  E  times  the  unit  strain. 

Thus  the  equation  is  the  analytical  expression  of  Hooke's 
law,  that  for  a  body  in  a  state  of  simple  strain  the  strain  is  pro- 
portional to  the  stress. 

The  longitudinal  strain  is  accompanied  by  an  alteration  in 

the  transverse  dimensions,  the  lateral  unit  strain  being  —  —  , 

m 

where  m  is  a  coefficient  which  usually  varies  from  3  to  4  for 
solid  bodies  and  is  approximately  4  for  the  metals  of  construc- 
tion. In  the  case  of  india-rubber,  if  the  deformation  is  small, 
m  is  about  2. 

Generally  the  deformation  may  be  calculated  per  unit  of 
original  length  without  sensible  error,  but  for  india-rubber  it  is 
more  accurate  to  make  the  calculation  per  unit  of  stretched 


length  {=  — — 

I  lateral  strain 

The   ratio  —  =  : : — -.- — : —  is  called  Poisson's  ratio. 

m       longitudinal  strain 

If  the  transverse  dimensions  of  a  bar  under  compression 
are  small  as  compared  with  the  length  L,  a  slight  disturbing 
force  will  cause  the  bar  to  bend  sideways,  and  the  bar  will  be 
subjected  to  a  bending  action  in  addition  to  the  compression. 
If  the  bar  is  to  be  capable  of  resisting  a  direct  thrust  only, 
the  ratio  of  L  to  its  least  transverse  dimension  should  not 
exceed  a  certain  limit  depending  upon  the  nature  of  the  ma- 
terial. For  example,  experiment  indicates  that  this  limit 
should  be  about  5  for  cast-iron,  10  for  wrought-iron,  7  for 
steel,  and  20  for  dry  timber. 

If  the  temperature  of  the  bar  is  raised  t°,  the  consequent 
strain  is  at,  a  being  the  coefficient  of  linear  dilatation ;  and  a 
stress  Eaf  will  be  developed  if  a  change  of  length  is  pre- 
vented. 


SPECIFIC    WEIGHT.— COEFFICIENT  OF  ELASTICITY.     143 

4.  Specific  Weight ;  Coefficient  of  Elasticity ;  Limit 
of  Elasticity  ;  Breaking  Stress. — Before  the  strength  of  a 
body  can  be  fully  known,  certain  physical  constants,  whose 
values  depend  upon  the  material,  must  be  determined. 

(a)  Specific  Weight. — The  specific  weight  is  the  weight  of  a 
unit  of  volume.     The  specific  weights  of  most  of  the  materials 
of  construction  have  been  carefully  found  and  tabulated.     If 
the  specific  weight  of  any  new  material  is  required,  a  conven- 
ient approximate  method  is  to  prepare  from  it  a  number  of 
regular  solids  of  determinate  volume  and  weigh  them   in  an 
ordinary  pair  of  scales.     The  ratio  of  the  total  weight  of  these 
solids  to  their  total  volume  is  the  specific  weight.     It  must  be 
remembered  that  the  weight  may  vary  considerably  with  time, 
etc. ;  thus  a  sample  of  greenheart  weighed  69.75  Ibs.  per  cubic 
foot  when  first  cut  out  of  the  log,  and  only  57  Ibs.  per  cubic 
foot  at  the  end  of  six  months.     When  the  strength  of  a  timber 
is  being  determined,  it  is  important  to   note  the   amount   of 
water  present  in  the  test-piece,  since  this  appears  to  have  a 
great  influence  upon  the  results. 

The  straining  of  a  structure  is  generally  largely  due  to  its 
own  weight. 

The  total  load  upon  a  structure  includes  all  the  external 
forces  applied  to  it,  and  in  practice  is  designated  dead  {perma- 
nent} or  live  (rolling),  according  as  the  forces  are  gradually  ap- 
plied and  steady,  or  suddenly  applied  and  accompanied  with 
vibrations.  For  example,  the  weight  of  a  bridge  is  a  dead 
load,  while  a  train  passing  over  it  is  a  live  load  ;  the  weight 
of  a  roof,  together  with  the  weight  of  any  snow  which  may 
have  accumulated  upon  it,  is  a  dead  load  ;  wind  causes  at  times 
excessive  vibrations  in  the  members  of  a  structure,  and  al- 
though often  treated  as  a  dead  load,  should  in  reality  be  con- 
sidered a  live  load. 

The  dead  loads  of  many  structures '(as  masonry  walls,  etc.) 
are  so  great  that  extra  or  accidental  loads  may  be  safely  disre- 
garded. In  cold  climates,  great  masses  of  snow  and  the  pene- 
trating effect  of  the  frost  necessitate  very  deep  foundations, 
which  proportionately  increase  the  dead  weight. 

(b)  Coefficient  of  Elasticity. — Generally  speaking,  a  knowl- 
edge of  the  external  forces  acting  upon  a  structure,  discloses 


144 


THEORY  OF  STRUCTURES. 


the  manner  of  their  distribution  amongst  its  various  members, 
but  the  deformation  of  these  members  can  only  be  estimated 
by  means  of  the  coefficient  of  elasticity,  which  expresses  the 
relation  between  a  stress  and  the  corresponding  strain. 

In  practice  it  is  usually  sufficient  to  assume  that  a  material 
is  elastic,  homogeneous,  and  isotropic,  and  its  deformation 
under  stress  may  be  found,  if  the  coefficients  of  elasticity,  of 
form,  and  of  volume  are  known. 

In  a  homogeneous  solid  there  may  be  twenty-one  distinct 
coefficients  of  elasticity,  which  are  usually  classified  under  the 
following  heads : 

(1)  Direct,   expressing   the   relation    between    longitudinal 
strains  and  normal  stresses  in  the  same  direction. 

(2)  Transverse,  expressing  the  relation  between  tangential 
stresses  and  strains  in  the  same  direction. 

(3)  Lateral,  expressing  the   relation   between   longitudinal 
strains  and  normal  stresses  at  right  angles  to  the  strains;  i.e.,  a 
lateral  resistance  to  deformation. 

(4)  Oblique,  expressing  other  relations  of  stress  and  strain. 
If  a  body  is  isotropic,  i.e.,  equally  elastic  in  all  directions, 

the  twenty-one  coefficients  reduce  to  two,  viz.,  the  coefficients 
of  direct  elasticity  and  of  lateral  elasticity.  Such  bodies,  how- 
ever, are  almost  wholly  ideal.  In  a  perfectly  elastic  body  E 
would  be  the  same  both  for  tension  and  compression.  In  the 

ordinary  materials  of  construction 
it  is  slightly  less  for  compression 
than  for  tension ;  but  if  the  stresses 
do  not  exceed  a  certain  limit  (§  (e\ 
page  145),  the  difference  is  so  slight 
that  it  may  be  disregarded. 

The  equation  f  —  E\  may  be 
represented  graphically  by  the 
straight  line  MON,  the  ordinate  at 
any  point  representing  the  unit 
stress  required  to  produce  the  unit 
strain  respresented  by  the  corresponding  abscissa. 

The  angle  MOY  =  tan  ~*E  =  tan  -'^V 

Coefficients  of  elasticity  must  be  determined  by  experiment. 


FlG-  l83- 


LIMIT  OF  ELASTICITY.  H5 

The  coefficients  of  direct  elasticity  for  the  different  metals 
and  timbers  are  sometimes  obtained  by  subjecting  bars  of  the 
material  to  forces  of  extension  or  compression,  or  by  observing 
the  deflections  of  beams  loaded  transversely.  The  coefficients 
for  blocks  of  stone  and  masonry  might  also  be  found  by  trans- 
verse loading ;  they  are  of  little,  if  any,  practical  use,  as,  on 
account  of  the  inherent  stiffness  of  masonry  structures,  their 
deformations,  or  settlings,  are  due  rather  to  defective  work- 
manship than  to  the  natural  play  of  elastic  forces. 

The  torsional  coefficient  of  elasticity,  i.e.,  the  coefficient  of 
elastic  resistance  to  torsion,  has  been  shown  by  experiment  to 
vary  from  two  fifths  to  three  eighths  of  the  coefficient  of  direct 
elasticity. 

(e)  Limit  of  Elasticity. — When  the  forces  which  strain  a 
body  fall  below  a  certain  limit,  the  body,  on  the  removal  of  the 
forces,  will  resume  its  original  form  and  dimensions  without 
sensible  change  (disregarding  any  effects  due  to  the  develop- 
ment of  heat)  and  may  be  treated  as  perfectly  elastic.  But  if 
the  forces  exceed  this  limit,  the  body  will  receive  a  permanent 
deformation,  or,  as  it  is  termed,  a  set. 

Such  a  limit  is  called  a  limit  of  elasticity,  and  is  the  greatest 
stress  that  can  be  applied  to  a  body  without  producing  in  it  an 
appreciable  and  permanent  deformation. 

This  is  an  unsatisfactory  definition,  as  a  body  passes  from 
the  elastic  to  the  non-elastic  state  by  such  imperceptible 
degrees  that  it  is  impossible  to  fix  any  exact  line  of  demarca- 
tion'between  the  two  states.  Fairbairn  defines  the  limit  mo're 
correctly,  as  the  stress  below  which  the  deformation  is  approxi- 
mately proportional  to  the  load  which  produces  it,  and  beyond 
which  the  deformation  increases  much  more  rapidly  than  the 
load.  In  fact,  both  the  elastic  and  ultimate  strengths  of  a  ma- 
terial depend  upon  the  nature  of  the  stresses  to  which  they  are 
subjected  and  upon  \k&  frequency  of  their  application.  For  ex- 
ample, in  experimenting  upon  bars  of  iron  having  an  ultimate 
tenacity  of  46,794  Ibs.  per  sq.  in.  and  a  ductility  of  20  %> 
Wohler  found  that  with  repeated  stresses  of  equal  intensity, 
but  alternately  tensile  and  compressive,  a  bar  failed  aft(br  56,430 
repetitions  when  the  intensity  was  33,000  Ibs.  per  sq.  in.  ;  a 


146  THEORY  OF  STRUCTURES. 

-second  bar  failed  only  after  19,187,000  repetitions  when  the 
intensity  was  18,700  Ibs.  per  sq.  in. ;  while  a  third  bar  remained 
intact  after  more  than  132,000,000  repetitions  when  the.  inten- 
sity was  16,690  Ibs.  per  sq.  in.  These  experiments  therefore 
indicated  that  the  limit  of  elasticity  for  the  iron  in  question, 
under  repeated  stresses  of  equal  intensity,  but  alternately 
'tensile  and  compressive,  lay  between  16,000  and  17,000  Ibs.  per 
sq.  in.,  which  is  much  less  than  the  limit  under  a  steadily  ap- 
plied stress.  Similar  results  have  been  shown  to  follow  when 
the  stresses  fluctuate  from  a  maximum  stress  to  a  minimum 
stress  of  the  same  kind. 

Generally  speaking,  then,  the  limit  of  elasticity  of  a  ma- 
terial subjected  to  repeated  stresses,  is  a  certain  maximum 
stress  below  which  the  condition  of  the  body  remains  unim- 
paired, 

Bauschinger's  experiments  indicate  that  the  application  to 
a  body  of  any  stress,  however  small,  produces  a  plastic  or 
permanent  deformation.  This,  perhaps,  is  sometimes  due  to  a 
want  of  uniformity  in  the  material,  or  to  the  bar  being  not 
quite  straight  initially.  In  any  case,  the  deformations  under 
loads  which  are  less  than  the  elastic  limit,  are  so  slight  as  to  be 
of  no  practical  account  and  may  be  safely  disregarded. 

The  main  object,  then,  of  the  theory  of  the  strength  of 
materials,  is  to  determine  whether  the  stresses  developed  in  any 
particular  member  of  a  structure  exceed  the  limit  of  elasticity. 
As  soon  as  they  do  so,  that  member  is  permanently  deformed, 
its  strength  is  impaired,  it  becomes  predisposed  to  rupture,  and 
the  safety  of  the  whole  structure  is  threatened.  Still,  it  must 
be  borne  in  mind  that  it  is  not  absolutely  true  that  a  material 
is  always  weakened  by  being  subjected  to  forces  superior  to 
this  limit.  In  the  manufacture  of  iron  bars,  for  instance,  each 
of  the  processes  through  which  the  metal  passes  changes  its 
elasticity  and  increases  its  strength.  Such  a  material  is  to  be 
treated  as  being  in  a  new  state  and  as  possessing  new  properties. 

The  strength  of  a  material  is  governed  by  its  tenacity  and 
rigidity,  and  the  essential  requirement  of  practice  is  a  tough 
material  with  a  high  elastic  limit. 

This  is  especially  necessary  for  bridges  and  all  structures 


BREAKING   STRESS.  147 

liable  to  constantly  repeated  loads,  for  it  is  found  that  these 
repetitions  lower  the  elastic  limit  and  diminish  the  strength. 

In  the  majority  of  cases,  experience  has  fixed  a  practical 
limit  for  the  stresses,  much  below  the  limit  of  elasticity.  This 
insures  greater  safety  and  provides  against  unforeseen  and 
accidental  loads,  which  may  exceed  the  practical  limit,  but 
which  do  no  harm  unless  they  pass  beyond  the  clastic  limit. 

Certain  operations  have  the  effect  of  raising  the  limit  of 
elasticity:  a  wrought-iron  bar  steadily  strained  almost  to  the 
point  of  its  ultimate  strength  and  then  released  from  strain  and 
allowed  to  rest,  experiences  an  elevation  both  of  tenacity  and 
of  the  elastic  limit. 

If  the  bar  is  stretched  until  it  breaks,  the  tensile  strength 
of  the  broken  pieces  is  greater  than  that  of  the  bar.  A  similar 
result  follows  in  the  various  processes  employed  in  the  manu- 
facture of  iron  and  steel  bars  and  wires :  the  wire  has  a  greater 
ultimate  strength  than  the  bar  from  which  it  was  drawn. 

Again,  iron  and  steel  bars,  subjected  to  long- continued  com- 
pression or  extension,  have  their  resistance  increased,  mainly 
because  time  is  allowed  for  the  molecules  of  the  metal  to  as- 
sume such  positions  as  will  enable  them  to  offer  the  maximum 
resistance  ;  the  increase  is  not  attended  by  any  ap- 
preciable change  of  density.  / 

Under  an  increasing  stress  a  brittle  material  will  / 

be  fractured  without  any  great  deformation,  while  a  / 

tough  material  will   become   plastic   and    undergo   a          J/ 
large  deformation.  ^P 

(d)  Breaking  Stress. — When    the    load    upon    a          / 
material    increases     indefinitely,    the    material    may          / 
merely  suffer  an  increasing  deformation,  but  generally         / 
a  limit   is   reached   at   which  fracture  suddenly  takes        / 
place.  / 

Cast-iron   is   perhaps   the    most    doubtful    of    all    / 
materials,  and  the  greatest  care  should  be  observed/ 
in  its  employment.     It    possesses   little   tenacity   or 
elasticity,  is  very  hard  and  brittle,  and  may  fail  sud-    FlG<  l84' 
denly  under  a  shock  or  an  extreme  variation  of  temperature. 
Unequal  cooling  may  predispose  the  metal  to  rupture,  and  its 


148 


THEORY  OF  STRUCTURES. 


strength  may  be  still  further  diminished  by  the  presence  of 
air-holes. 

Cast-iron  and  similar  materials  receive  a  sensible  set  even 
under  a  small  load,  and  the  set  increases  with  the  load.  Thus 
at  no  point  will  the  stress-strain  curve  be  absolutely  straight, 
and  the  point  of  fracture  will  be  reached  without  any  great 
change  in  the  slope  of  the  curve  and  without  the  development 
of  much  plasticity. 

Wrought-iron  and  steel  are  far  more  uniform  in  their  be- 
havior, and  obey  with  tolerable  regularity  certain  theoretical 
laws.  They  are  tenacious,  ductile,  have  great  compressive 
strength,  and  are  most  reliable  for  structural  purposes.  Their 
strength  and  elasticity  may  be  considerably  reduced  by  high 
temperatures  or  severe  cold. 

When  a  bar  of  such  material  is  tested,  the  stress-strain 
jpurve  (/=  ±  Ety,  as  has  already  been  pointed  out,  is  almost 
absolutely  straight  within  the  elastic  limit,  e.g.,  from  O  to  A  in 

tension  and  from  O  to  B  in  com- 
pression. As  the  load  increases 
beyond  the  elastic  limit,  the  in- 
creasing deformation  becomes 
plastic  and  permanent,  and  the 
stress-strain  diagram  takes  an  ap- 
preciable curvature  between  the 
limits  A  and  B  and  the  points  D 
and  E  corresponding  to  the  maxi- 
mum loads.  In  tension,  as  soon 
as  the  point  D  is  reached,  the  bar 
rapidly  elongates  and  is  no  longer 
able  to  sustain  the  maximum  load, 
its  sectional  area  rapidly  dimin- 
ishes, and  fracture  ultimately  takes 
place  under  a  load  much  less  than 
the  maximum  load.  The  point  of 
fracture  is  represented  in  the  figure 
by  the  point  F  the  ordinate  of  F  being  the  actual  ultimate 

final  load  on  the  bar 
intensity  of  stress  =  —  section  • 


FIG.  185. 


BREAKING   STRESS.  149 

The  exact  form  of  the  stress-strain  curve  between  D  and  F 
is  unknown,  as  no  definite  relation  has  been  found  to  exist  be- 
tween the  stress  and  strain  during  the  elongation  from  D  to  F. 

Ordinarily,  the  breaking  tensile  stress  has  been  defined  to  be 
the  maximum  load  applied  divided  by  the  initial  sectional  area 
of  the  bar ;  but  this,  although  convenient,  is  manifestly  in- 
correct. 

It  is  important  to  note  that,  as  the  deformation  gradually 
increases  under  the  increasing  load,  the  molecules  of  the  ma- 
terial require  greater  or  less  time  to  adjust  themselves  to  the 
new  condition. 

During  the  tensile  test  of  a  ductile  material  there  is,  at 
some   point    beyond    the  elastic   limit,   an 
abrupt  break  GH  in  the  continuity  of  the 
stress-strain  curve,  the  curve  again  becom- 
ing continuous  from  H  to  D. 

The  point  G  has  been  called  the  Yield 
Point  or  the  Breaking-down  Point,  and  the 
deformation  from  H  onward  is  almost 
wholly  plastic  or  permanent. 

In  compression  there  is  no  local  stretch 
as  in  tension,  and  there  is  consequently 
no  considerable  change  in  the  curvature  of  FIG.  iS6. 

the  compression  stress-strain  curve  up  to  the  point  of  fracture. 

Timber  is  usually  tested  by  being  subjected  to  the  action 
of  tensile,  compressive,  or  transverse  loads.  Other  character- 
istics, however,  must  be  known  before  a  full  conception  of  the 
strength  of  the  wood  can  be  obtained.  Thus  the  specific 
weight  must  be  found  ;  the  amount  of  water  present,  the  loss 
in  drying,  and  the  corresponding  shrinkage  should  be  deter- 
mined ;  the  structural  differences  of  the  several  specimens,  the 
rate  of  growth,  etc.,  should  be  observed. 

The  chief  object  of  experiments  upon  masonry  and  brick- 
work is  to  discover  their  resistance  to  compression,  i.e.,  their 
crushing  strength.  In  fact,  their  stiffness  is  so  great  that  they 
may  be  compressed  up  to  the  point  of  fracture  without  sensible 
change  of  form,  and  it  is  therefore  very  difficult,  if  not  impos- 
sible, to  observe  the  limit  of  elasticity. 


1 50  THEORY  OF  STRUCTURES. 

The  cement  or  mortar  uniting  the  stones  and  bricks  is  most 
irregular  in  quality.  In  every  important  work  it  should  be  an 
invariable  rule  to  prepare  specimens  for  testing.  The  crushing 
strength  of  cement  and  of  mortar  is  much  greater  than  the  ten- 
sile strength,  the  latter  being  often  exceedingly  small.  Hence  it 
is  advisable  to  avoid  tensile  stresses  within  a  mass  of  masonry, 
as  they  tend  to  open  the  joints  and  separate  the  stones  from 
one  another.  Attempts  are  frequently  made  to  strengthen 
masonry  and  brickwork  walls  by  inserting  in  the  joints  tarred 
and  sanded  strips  of  hoop-iron.  Their  utility  is  doubtful,  for, 
unless  well  protected  from  the  atmosphere,  they  oxidize,  to  the 
detriment  of  the  surrounding  material,  and,  besides  this,  they 
prevent  an  equable  distribution  of  pressure.  They  are,  how- 
ever, far  preferable  to  bond-timbers. 

The  working  load  (or  stress,  or  strength)  is  the  maximum 
stress  which  a  material  can  safely  bear  in  ordinary  practice, 
and  depends  both  upon  the  character  (see  Art.  5, .below)  of  the 
stress  and  upon  the  ultimate  strength  of  the  material,  the  ratio 
of  the  ultimate  or  breaking  stress  to  the  working  stress  being 
usually  called  a  factor  of  safety.  For  example,  the  factor  is 
about 

3  for  long-span  iron  bridges,  or  bridges  having  great  weight 
as  compared  with  the  live  load  (a  moving  train). 

4  for  ordinary  iron  bridges. 

5  for  ordinary  metal  shafting. 

8,  10,  and  even  more  for  long  struts  and  members  subjected 
to  repeated  stresses  of  varying  magnitude. 

10  is  also  generally  taken  to  be  the  factor  of  safety  for 
timber. 

Under  a  steady,  or  a  merely  statical  load,  even  as  great  as 
•J-  of  the  breaking  stress,  a  member  of  a  structure  may  prob- 
ably not  be  unsafe. 

5.  Wohler's  Law. — It  is  now  generally  admitted  that 
variable  forces,  constantly  repeated  loads,  and  continued  vibra- 
tions diminish  the  strength  of  a  material,  whether  they  pro- 
duce stresses  approximating  to  the  elastic  limit,  or  exceedingly 
small  stresses  occurring  with  great  rapidity.  Indeed  many 
engineers  design  structures  in  such  a  manner,  that  the  several 


WOHLER'S  LAW.  !$! 

members  are  strained  in  one  way  only,  so  convinced  are  they 
of  the  evil  effect  of  alternating  tensile  and  compressive  stresses. 
Although  the  fact  of  a  variable  ultimate  strength  had  thus  been 
tacitly  acknowledged  and  often  allowed  for,  Wohler  was  the 
first  to  give  formal  expression  to  it,  and,  as  a  result  of  obser- 
vation and  experiment,  enunciated  the  following  law: 

"  That  if  a  stress  t,  due  to  a  static  load,  cause  the  fracture 
of  a  bar,  the  bar  may  also  be  fractured  by  a  series  of  often-re- 
peated stresses,  each  of  which  is  less  than  t\  and  that,  as  the 
differences  of  stress  increase,  the  cohesion  of  the  material  is 
affected  in  such  a  manner  that  the  minimum  stress  required  to 
produce  fracture  is  diminished." 

This  law  is  manifestly  incomplete.  In  Wohler's  experi- 
ments the  applications  of  the  load  followed  each  other  with 
great  rapidity,  yet  a  certain  length  of  time  was  required  for  the 
resulting  stresses  to  attain  their  full  intensity  ;  the  influence  due 
to  the  rapidity  of  application,  to  the  rate  of  increase  of  the 
stress, -and  to  the  duration  of  individual  strains  still  remains  a 
subject  for  investigation. 

The  experiments,  however,  show  that  the  rate  of  increase  of 
repetitions  of  stress  required  to  produce  fracture,  is  much  more 
rapid  than  the  rate  of  decrease  of  the  stresses  themselves,  and 
depends  both  upon  the  maximum  stress  and  upon  the  differ- 
ence or  fluctuation  of  stress. 

The  effect  of  repeated  stresses  of  equal  intensity,  but  alter- 
nately tensile  and  compressive,  has  been  already  pointed  out 
in  Art.  4. 

Bars  of  the  same  material  repeatedly  bent  in  one  direction, 
bore  31,132  Ibs.  per  square  inch  when  the  load  was  wholly  re- 
moved between  each  bending,  and  45,734  Ibs.  per  square  inch 
when  the  stress  fluctuated  between  45,733  Ibs.  and  24,941  Ibs. 

The  table  on  page  152  gives  the  results  of  similar  experi- 
ments on  steel. 

The  axle-steel  was  found  to  bear  22,830  Ibs.  per  square  inch, 
when  subjected  to  repeated  shears  of  equal  intensity  but  oppo- 
site in  kind,  and  29,440  Ibs.  per  square  inch,  when  the  shears 
were  of  the  same  kind.  It  would  therefore  appear  that  the 
shearing  strengths  of  the  metal  in  the  two  cases  are  about  £ 


152 


THEORY  OF  STRUCTURES. 


of  the  strengths  of  the  same  metal  under  alternate  bending  and 
under  bending  in  one  direction,  respectively. 


Character  of  Fluctuation. 

Maximum  Resistance  to  Repeated 
Stresses  in  Ibs.  per  square  inch. 

Axle-steel. 

Spring-steel  (un- 
hardened). 

Alternating  stresses  of  equal  intensity  .... 
Complete  relief  from  stress  between    each 

29,000,  —  29,000 
49,890,           O 
83,110,      36,380 

52,000,        o 

93,500,    62,240 

Partial    relief   from    stress    between    each 

From  torsion  experiments  with  various  qualities  of  steel, 
the  important  result  was  deduced,  that  the  maximum  resistance 
of  the  steel  to  alternate  twisting  was  |-  of  the  maximum  resist- 
ance of  the  same  steel  to  alternate  bending. 

Wohler  proposed  2  as  a  factor  of  safety,  and  considered 
that  the  maximum  permissible  working  stresses  should  be  in 
the  ratios  of  1:2:3,  according  as  members  are  subjected  to 
alternate  tensions  and  compressions  (alternate  bending),  to 
tensions  alternating  with  entire  relief,  or  to  a  steady  load. 

The  weakening  of  metal  by  repeated  stresses  has  been 
called  fatigue,  and  is  much  more  injurious  to  iron  and  steel 
under  tension  than  under  compression.  Egleston's  investiga- 
tions have  shown  that  a  fatigued  metal  may  sometimes  be 
restored  by  rest  or  by  annealing. 

From  the  law,  however,  as  it  stands  formulae  may  be  de- 
duced which,  it  is  claimed,  are  more  in  accordance  with  the 
results  of  experiment,  give  smaller  errors,  and  insure  greater 
safety  than  the  false  assumption  of  a  constant  ultimate 
strength. 

The  formulae  necessarily  depend  upon  certain  experimental 
results,  but  in  applying  them  to  any  particular  case,  it  must  be 
remembered  that  only  such  results  should  be  employed,  as 
have  been  obtained  for  material  of  the  same  kind  and  under 
the  same  conditions  as  the  material  under  consideration.  The 
effects  due  to  faulty  material,  rust,  etc.,  are  altogther  indeter- 
minate, so  that  no  formula  can  be  perfectly  universal  in  its 
application.  Hence  the  necessity  for  factors  of  safety,  with 
values  depending  upon  the  class  of  structure,  still  exists. 


LAUNHARDT'S  FORMULA.  I  53 

A  brief  description  of  the  principal  of  these  formulae  will 
now  be  given,  and  in  the  discussion 

/,  the  statical  breaking  strength,  is  the  resistance  to  fracture 
under  a  static  or  under  a  very  gradually  applied  load. 

?/,  the  primitive  strength,  is  the  resistance  to  fracture  under 
a  given  number  of  repeated  stresses,  the  stress  in  each  repeti- 
tion remaining  unchanged  in  kind,  i.e.,  being  due  either  to  a 
tension,  a  compression,  or  a  shear. 

s,  the  vibration  strength,  is  the  resistance  to  fracture  under 
alternating  stresses  of  equal  intensities,  but  different  in  kind, 
due  to  a  vibratory  motion  about  the  unstrained  state  of  equi- 
librium. 

b  is  the  admissible  stress  per  unit  of  sectional  area 

/MS  the  effective  sectional  area  and  is 

numericallv  absolute  maximum  load 


6.  Launhardt's  Formula. — A  bar  of  unit  sectional  area  is 
subjected  to  stresses  (B)  which  are  either  wholly  tensile,  wholly 
compressive,  or  wholly  shearing,  and  which  vary  from  a  maxi- 
mum #,  (  =  max.  B)  to  a  minimum  #2  (=  min.  If). 

Let  <7,  —  <?Q  —  d  —  the  maximum  difference  .of  stress. 

T  a,.         min.  B 

Let  -2  = -=  =  0. 

al        max.  B 

If  a.2  =  o,  #,  =  d  —  u. 

I f  d  —  o,  al  —  a^—t. 

By  Wohkr's  law, 

*,cc  <*=/</, (i) 

/being  an  unknown  coefficient  of  which  the  value  remains  to 
be  determined. 

If  d  —  o,  al  =  t     and    /  =   oo. 

If  d  —  u,  #,  =  d    and    f  =.  i. 


154  THEORY   OF  STRUCTURES. 

Launhardt's  assumption,  viz.,f=  —  —  ,  satisfies  these  ex- 

/       a1 

treme  conditions,  and  also  gives  intermediate  values  of  al  which 
closely  agree  with  the  results  of  the  most  reliable  experiments. 
Hence  (i)  becomes 

t—H          t-u 


and 

/         t  —  u  a2]         f         *  —  u\  ,  ^ 

.-.  a  .=  u  I  -\  ---  }=u(i-\  --  0.     .     .     (2) 
\  u      aj          \  n    ml 

This  is  Launhardt's  formula,  and  is  an  analytical  expression 
of  Wohler's  Law. 

Wohler  in  his  bending  experiments  upon  Phcenix  axle-iron 
found  that  u  =  2195*  per  cent.2*  and  t  =  4020*  per  cent.1; 

/  —  u  _    ^ 
~~u  6' 

The  same  iron  under  tension  gave  u  =  2195*  per  cent.3  and 
/  =  3290*  per  cent.2  ; 

t  —  U          ! 


U  2 

Choosing  the  most  unfavorable  case,  and,  in  order  to  insure 
greater  safety,  taking  u  =  2100*  per  cent.2,  equation  (2)  becomes 

a,  =  2100(1+-)  .......     (3) 

If  3  is  the  factor  of  safety, 

6  =  700(1  +  |).  .......    (4) 


*  k  per  cent.'2  is  an  abbreviation  for  kilogrammes  per  square  centimetre. 
One  kilogramme  per  square  centimetre  is  equivalent  to  14.2232  Ibs.  per  sq.  in. 


LAUNHARDT'S  FORMULA.  155 

In  his  bending  experiments  upon  Krupp  cast-steel  (untem- 
pered)  it  was  found  that  u  =  3510*  per  cent.2  and  t  =  7340*  per 
cent2.; 

t  —  u  __  7 
~~u          6* 

But  steel  varies  considerably  in  strength,  and  great  care 
must  be  exercised  in  its  use,  especially  in  bridge  construction. 
For  this  reason  take  u  —  3300*  per  cent.2  and  t  =  6oook  per  cent.2; 

/  —  u  _     9 
~H~    =T? 

and  (2)  becomes 

«,  =  33oo(i  +  £0) (5) 

If  3  is  the  factor  of  safety, 

^=  1100(1+^-0) (6) 

EXAMPLE  I. — The  stresses  upon  a  bar  of  Phoenix  axle-iron, 
normal  to  its  cross-section,  vary  from  a  maximum  tension  of 
50000*  to  a  minimum  tension  of  20000*.  Determine  the  admis- 
sible stress  per  cent.2  and  the  necessary  sectional  area 

By  (4), 

f  I  20000  \ 

b  =  700(1  +  2^^-J  =  840*  Per  cent1, 
and 

50000        50000 
.'.  F  =±  — T—  =  -Q— -  —  59-52  sq.  centimetres. 

u  040 

Let/  be  the  dead  load  and  q  the  total  load,  per  lineal  unit 
of  length,  upon  the  flanges  of  roof  and  bridge  trusses. 


THEORY  OF  STRUCTURES. 

.-.  0  =  —  ,  and  equations  (4)  and  (6)  respectively  become 
*=    700(1+-)  .......    (7) 


Ex.   2. — Determine  the  limiting  stress  per  cent.2  for  the 
flanges  of  a  wrought-iron  lattice  girder  when  the  ratio  of  the 

dead  load  to  the  greatest  total  load  is  —=-. 


By  (7), 

b  =  700(1 +~)  =  800*. 

»  ^   s*>* 


7.  Weyrauch's  Formula. — Let  a  bar  of  a  unit  sectional 
area  be  subjected  to  stresses  which  are  alternately  different  in 
kind,  and  which  vary  from  an  absolute  numerical  maximum  a' 
(=  max.  B)  of  the  one  kind  to  a  maximum  a"  (=  max.  B')  of 
the  other  kind. 

Let  a'  -\-  a"  =  d  =  the  maximum  numerical  difference  of 
stress. 

a"        max.  B' 

Let 


If  a"  =0,  a'  =  d  =  u. 

T    £  f  f       /       ^ __ 

By  Wohler's  Law, 

a1  a  </  =//, (9) 

^"  being  an  unknown  coefficient  of  which  the  value  remains  to 
be  determined. 

If  a'  =  u,  f  —  I. 

Tf  /»'  nr    C  /•=  i 


WEYRAUCH'S  FORMULA.  157 

Weyrauch's   assumption,   viz.,  f  =  -—  —  -,  ,   satisfies 

226   —  —   S  ~~~  d 

these  extreme  conditions,  the  most  reliable  results  of  the  few 
experiments  yet  recorded,  and  also  Wohler's  deduction  that  a' 
diminishes  as  d  increases  and  vice  versa. 
Hence  (9)  becomes 


and 

u-s 


This  is  Weyrauch's  formula,  and  it  may  be  always  applied 
to  those  cases  in  which  a  member  is  subjected  to  stresses  alter- 
nating between  tension*  and  compression,  or  due  to  shearing 
actions  in  opposite  directions. 

In  the  Phcenix  iron  experiments  already  referred  to  it  was 
found  that  s  =  1170*  per  cent.2  ; 

u  —  s        7 
u       ~  15" 

Taking  u  =  2100*  as  before,  and  making  -  =  —  ,  (10) 
becomes 


(II) 

If  3  is  the  factor  of  safety, 

*  =  700(1  --£)  .......     (12) 


Weyrauch  considers  3  to  be  the  proper  factor  of  safety  for 
bridges  and  similar  structures.  It  is  also  a  suitable  factor  for 
the  parts  of  machines  subjected  to  determinate  straining 
actions.  A  larger  factor  will  be  required  when  other  con- 
tingencies have  to  be  provided  against. 


158  THEORY  OF  STRUCTURES. 

In  the  steel  experiments,  Wohler  found  that  s  =  2050*  per 
cent.3  ; 

u-s  =  _5_ 

U  12  • 

Taking  u  =  3300*  and  s  —  1800*, 

«  -  J      _5 

»       '11' 
and  (10)  becomes 

^  =  3300(1-^00  .......    (13) 

If  3  is  the  factor  of  safety, 

b=  1100(1  --T5T0')  ....... 


If  a  very  soft  steel  is  employed  in  the  construction  of  a 
bridge,  it  may  be  advisable  to  diminish  still  further  the  ad- 
missible stress  per  unit  of  sectional  area.  For  example,  it  may 
be  assumed  that  t  —  5200*,  u  =  3000*,  and  s  =  1500*,  so  that 
(2)  and  (10)  respectively  become 

a,  =  3000(1  +|0)    ......     (15) 

and 

a'  =  3000(1  -  £00.       .....     (16) 

EXAMPLE.  —  The  stresses  in  a  wrought-iron  bar  normal 
to  its  cross-section,  vary  between  a  tension  of  40000*  and  a 
compression  of  30000*.  Find  the  sectional  area  (disregarding 
buckling). 

By  (12) 

b  =  700(1  -  J.  X  t*W  =  437-5"  per  cent.9. 


40000 

/.  F=  -      -  =  91.42  sq.  centimetres. 
437-5 

Shearing  Stresses.  —  For  shearing  stresses  in  opposite  direc- 
tions Wohler  found,  in  the  case  of  Krupp  cast-steel  (untem- 


WEYRAUCH'S  FORMULA.  1 59 

pered),  that  u  =  2780*  per  cent.2  and  s  =  1610*  per  cent.2,  or 
about  •§-  of  the  corresponding  values  for  stresses  which  are 
alternately  tensile  and  compressive,  and  it  may  be  generally 
assumed,  that  the  value  of  b  for  shearing  stresses,  is  £  of  its 
value  for  stresses  which  are  alternately  tensile  and  compressive, 
and  which  have  the  same  ratio  0'. 

8.  Unwin  has  proposed  to  include  all  cases  of  fluctuating 
stress  in  the  formula 


a  being  the  actual  strength,  d  the  fluctuation  of  stress,  /  the 
statical  breaking  strength,  and  n  a  coefficient  whose  value 
remains  to  be  determined. 

When  d  =  o,  the  load  is  steady  and  a'  =  t. 

When  d  =  a',  the  load  alternates  with  entire  relief  and 


a'  —  2t(Vi  -\-tf-  n). 

When  d  =  20! ',  the  stresses  are  alternately  tensile  and  com- 
pressive and  of  equal  intensity.  The  stress  fluctuates  from 

a!  to  —  d ',  and  a'  =  — . 
2n 

In  these  extreme  cases,  if  n  is  made  equal  to  1.42  for 
wrought-iron  and  to  1.66  for  steel,  results  are  obtained  almost 
identical  with  those  given  in  Arts.  6  and  7.  The  formula  may 
therefore  be  assumed  to  be  approximately  correct  for  inter- 
mediate cases. 

The  mean  value  of  n  for  iron  and  steel  seems  to  be  f,  so 
that  the  formula  may  be  written 


EXAMPLE. — One  of  the  diagonals  of  a  bowstring  truss  has 
a  sectional  area  of  3  square  inches,  and  is  subjected  to  stresses 


l6o  THEORY  OF  STRUCTURES. 

which  fluctuate  between  a  tension  of  14  tons  and  a  compression 
of  6  tons.     Find  the  statical  strength  of  the  iron. 


14  —  (—6)       20 
d  =  fluctuation  of  stress  =  —  -  *  --  -  =  — 


t  =  10.17  tons  per  sq.  in. 

9.  Remarks  upon  the  Values  of  f,  u,  s,  and  6. — As  yet 

the  value  of  u  in  compression  has  not  been  satisfactorily  deter- 
mined, and  for  the  present  its  value  may  be  assumed  to  be  the 
same  both  in  tension  and  compression. 

If,  as  Wohler  states,  "repeated  stresses"  are  detrimental  to 
the  strength  of  a  material,  then  the  values  of  u  and  s  diminish 
as  the  repetitions  increase  in  number,  and  are  minima  in  struc- 
tures designed  for  a  practically  unlimited  life. 

Only  a  very  few  of  Wohler's  experiments  give  the  values  of 
t,  u,  s,  and  a,  so  that  Launhardt's  and  Weyrauch's  assumptions 
for  the  value  of /must  be  regarded  as  tentative  only,  and  re- 
quire to  be  verified  by  further  experiments.  The  close  agree- 
ment of  Wohler's  results  from  tests  upon  untempered  cast-steel 
(Krupp),  with  those  given  by  Launhardt's  formula,  may  be  seen 
from  the  following: 

For  /=  iioo  centners'*  per  sq.  zoll,  Wohler  found  that 
u  =  500  centners  per  sq.  zoll.  Thus  (2)  becomes 

«,  = 

and 

.'.  <2,a  —  SOOa,  —  6ootf2  —  o. 

Hence  for       a9  =    o,       250,     400,     600,     iioo, 
Launhardt's  formula  gives 

^  —  500,     710,     800,     900,     iioo; 

*  A  centner  =  110.23  pounds.     A  square  zoll  =  1.0603  square  inches. 


REMARKS    UPON    THE   VALUES   OF  t,  U,  S,  AND  b.       1 6 1 

while  Wohler's  experiments  gave 

a,  =  500,     700,     800,     900,     1 100. 

Again,  with  Phoenix  iron,  for  t  =  500  centners  per  sq.  zoll, 
u  was  found  to  be  300  centners  per  sq.  zoll,  and 


or 


a?  —  300^  —  250^  —  o. 


If  a^  =  240,  tfj  —  436.8,  which  almost  exactly  agrees  with 
the  result  given  by  the  tension  experiments, 

In  general,  the  admissible  stress  per  square  unit  of  sectional 
area  may  be  expressed  in  the  form 

b  =  v(i  ±  m<j>),      ......     (17} 

v  and  m  being  certain  coefficients  which  depend  upon  the 
nature  of  the  material  and  also  upon  the  manner  of  the  loading. 
Consider  three  cases,  the  material  in  each  case  being  wrought- 
iron  : 

(a)  Let  the  stresses  vary  between  a  maximum  tension  and 
an  equal  maximum  compression  ;  then 

0=1, 
and 

.•.  b  =  700(1  —  i)  —  350*  per  cent.2. 

(b)  Let  the  material  be  subjected    to   stresses  which    are 
either  tensile  or  compressive,  and  let  it  always  return  to  the 
original  unstrained  condition  ;   then 

min.  B  —  o,     or     max.  B'  =  o,     and     .*.  0  =  o. 
.-.   b  =  700(1  ±  o)  =  700*  per  cent.2. 

(c)  Let  the  material  be  continually  subjected  to  the  same 
dead  load ;  then 

min.  B  —  max.  B 


1 62  THEORY  OF  STRUCTURES. 

and 

..-.  b  =.  700(1  -|-  J)  =  1050*  per  cent.'  =  14,934  Ibs.  per  sq.  in., 

-which  is  one  third  of  the  ultimate  breaking  strength,  viz., 
1050'  per  cent.2. 

Thus  in  these  three  cases  the  admissible  stresses  are  in  the 
-.ratios  of  1:2:3,  ratios  which  have  been  already  adopted  in  ma- 
chine construction  as  the  result  of  experience. 

Wohler,  from  his  experiments  upon  untempered  cast-steel 
(Krupp),  concluded  that  for  alternations  between  an  unloaded 
condition  and  either  a  tension  or  a  compression,  b  =  iioo,  and 
for  alternations  between  equal  compressive  and  tensile  stresses, 
b  =  580. 

In  America  it  has  often  been  the  practice  to  take 

max.  B  +  max.  B'       a'  +  a" 

_£~*    " — "    


700  700 

for  stresses  alternately  tensile  and  compressive,  it  being  as- 
sumed that  if  the  stresses  are  tensile  only,  their  admissible 
values  may  vary  from  o*  to  700*  per  cent.8. 

a  ?ooF  a'         700 

Smce0    =r,.'.*  ==>'  and  •'•*  =      =  >-    (l8) 


Comparing  this  with  (12), 

for  0'  =    o,         i,         J,         j,          I, 
(18)  gives^  =  700,     560,     467,     400,     350, 
and  (12)  gives  b  —  700,     612,     525,     437,     350. 

10.  Flow  of  Solids.  —  When  a  ductile  body  is  strained 
beyond  the  elastic  limit,  it  approaches  a  purely  plastic  con- 
dition in  which  a  sufficiently  great  force  will  deform  the  body 
indefinitely.  Under  such  a  force,  the  elasticity  disappears  and 
the  material  is  said  to  be  in  a  fluid  state,  behaving  precisely 
like  a  fluid.  For  example,  it  flows  through  orifices  and  shows 
a  contracted  section.  The  stress  developed  in  the  material  is 
called  the  fluid  pressure  or  coefficient  of  fluidity. 

The  general  principle  of  the  flow  of  solids,  deduced  by 
Tresca,  may  be  enunciated  as  follows: 


FLO  W  OF  SOLIDS.  163 

A  pressure  upon  a  solid  body  creates  a  tendency  to  the  relative 
motion  of  the  particles,  in  the  direction  of  least  resistance. 

This  gives  an  explanation  of  the  various  effects  produced 
in  materials  by  the  operations  of  wire-drawing,  punching,  shear- 
ing, rolling,  etc.,  and  in  the  manufacture  of  lead  pipes.  Prob- 
ably it  also  explains  the  anomalous  behavior  of  solids  under 
certain  extreme  conditions. 

Rails  which  have  been  in  use  for  some  time  are  found  to 
have  acquired  an  elongated  lip  at  the  edge.  This  is  doubtless 
due  to  the  flow  of  the  metal  under  the  great  pressures  to  which 
the  rails  are  continually  subjected.  Other  examples  of  the  flow 
of  solids  are  to  be  observed  in  the  contraction  of  stretched  bars 
and  in  the  swelling  of  blocks  under  compression.  The  period 
of  fluidity  is  greater  for  the  more  ductile  materials,  and  may- 
disappear  altogether  for  certain  vitreous  and  brittle  substances. 

In  punching  a  piece  of  wrought-iron  or  steel,  the  metal  is 
at  first  compressed  and  flows  inwards,  while  the  shearing  only 
commences  when  the  opposite  surface  begins  to  open.  A  case 
brought  under  the  notice  of  the  author  may  be  mentioned  in 
illustration  of  this.  The  thickness  of  a  cold-punched  nut  was 
1.75  inches,  the  nut-hole  was  .3125  inch  in  diameter,  and  the 
length  of  the  piece  punched  out  was  only  .75  inch.  Thus  the 
flow  must  have  taken  place  through  a  depth  of  I  inch,  and  the 
shearing  through  a  depth  of  .75  inch.  Hence  the  surface 
really  shorn  was  TT  x  .3125  X  «75  =  -73^  sq.  in.  in  area,  and  a 
measure  of  the  shearing  action  is  the  product  of  this  surface 
area  and  the  fluid  pressure.  The  nature  of  the  flow  may  be 
observed  by  splitting  a  cold  punched  nut  in  half  and  treating 
the  fractured  surfaces  with  acid,  after  having  planed  them  and 
given  them  a  bright  polish.  The  metal  bordering  the  core  will 
be  found  curved  downwards,  the  curvature  increasing  from  the 
bottom  to  the  top,  and  well-defined  curves  will  mark  the  sepa- 
rating planes  of  the  plates  which  were  originally  used  in  piling 
and  rolling  the  iron. 

Jn  experimenting  upon  lead,  Tresca  placed  a  number  of 
plates,  one  above  the  other,  in  a  strong  cylinder,  Fig.  188,  page 
165,  with  a  hole  in  the  bottom.  Upon  applying  pressure  the 
lead  was  always  found  to  flow  when  the  coefficient  of  fluidity 


164  THEORY  OF  STRUCTURES. 

was  about  2844  Ibs.  per  sq.  in.,  the  difference  of  stress  being 
double  this  amount.  The  separating  planes  assumed  curved 
forms  analogous  to  the  corresponding  surfaces  of  flow  when 
water  is  substituted  in  the  cylinder  for  the  lead. 

The  flow  of  ductile  metals,  e.g.,  copper,  lead,  wrought-iron, 
and  soft  steel,  commences  as  soon  as  the  elastic  limit  is  ex- 
ceeded, and  in  order  that  the  flow  may  be  continuous  the  dis- 
torting stress  must  constantly  increase.  On  the  other  hand, 
in  the  case  of  truly  plastic  bodies,  flow  commences  and  con- 
tinues under  the  same  constant  stress.  It  evidently  depends 
upon  the  hardness  of  the  material,  and  has  been  called  the  co- 
efficient of  hardness.  The  longer  the  stress  acts  the  greater  is 
the  deformation,  which  gradually  increases  indefinitely  or  at  a 
diminishing  rate. 

Experiment  shows  that  there  is  very  little  alteration  in  the 
density  of  a  ductile  body  during  its  plastic  deformation,  and 
Tresca's  analytical  investigations  are  based  on  the  assumption 
that  the  body  is  deformed  without  sensible  change  of  volume. 

Consider  a  prismatic  bar  undergoing  plastic  deformation. 

Let  L  be  the  length  and  A  the  section  of  the  bar  at  com- 
mencement of  deformation. 

Let  L  -f-  x  be  the  length  and  a  the  section  of  the  bar  at  a 
subsequent  period. 

Let/  be  the  intensity  of  the  fluid  pressure. 

Since  the  volume  remains  unchanged, 

LA  =  (L  ±  x}a,  ......     .     (i) 

the  positive  or  negative  sign  being  taken  according  as  the  ba*- 
is  in  tension  or  compression. 

Let  Pl  be  initial  force  on  bar. 

Let  P  be  force  on  bar  when  its  length  is  L  ±  x.    Then 


P        a  L 

=    = 


Hence  P(L  ±  x)  =  P,L  =  a  constant,  ....     (3) 


FLOW  OF  SOLIDS. 


I65 


FIG.  187. 


and  the  force  diminishes  as  the  bar  stretches  and  increases  ^s 

the  bar  contracts  under  pressure. 
If  equation  (3)  be  referred  to  rect- 
angular axes,  the  ordinates  repre- 
senting different  values  of  P  and 
the  abscissae  the  corresponding 
values  of  x,  the  stress-strain  dia- 
grams,  tt  in  tension  and  cc  in  com- 
pression, are  hyperbolic  curves, 
having  as  asymptotes  the  axis  of 
x,  XOX,  and  a  line  parallel  to  the 
axis  of  y  at  a  distance  from  it 
equal  to  the  length  L  of  the  bar. 

Next  consider  a  metallic  mass 
(e.g.,  lead)   resting  upon  the  end 

CD  of  a  cylinder  of  radius  R,  and  filling  up  a  space  of  depth 

D.     A  hole  of  radius  r  is  made  at  the  centre 

of   the    face  CD,  through   which    the    mass 

flows  under  the  pressure  of  fluidity  exerted 

by  a  piston.     When  the  mass  has  been  com- 

pressed  to  the  thickness  DO  —  x,  let  y  be 

the  corresponding  length  KE  of  the  "jet." 
First,  assume  that  the  specific  weight  of 

the  mass  remains  constant. 

If  dx  be  the  diminution  in  the  thickness 

DO  corresponding  to  an  increase  dy  in  the 

length  of  the  jet,  then 

TtR'dx  -f-  rtfdy  —  O.    .     .     (i) 

Integrating  eq.  I,  and  remembering  that 
y  —  o  when  x  —  D, 


R\D  -x]-  ry  =  o 


(2) 


Second,  assume  that  the  cylindrical  portion  EFGH  is  gradu- 
ally transformed  into  NMPLKQN,  of  which  the  part  PMNQ 
is  cylindrical,  while  the  diameter  of  the  part  PLKQ  gradually 


1 66  THEORY  OF  STRUCTURES. 

increases  from  the  face  of  the  cylinder  to  KL  (  =  EF\  at  the 
end  of  the  jet.     Then 


r*)dx=  amount    of   metal   which   flows    into   the 
central  cylinder 

=  znrdrx, (3) 


dr  being  the  depth  to  which  the  metal  penetrates. 

Third,  assume  that  the  diminution  of  the  diameter  of  the 
cylindrical  portion  PMNQ  is  directly  proportional  to  the  said 
diameter. 

Then,  if  2  be  the  radius  of  the  cylinder  PQNM, 


dr       dz 

7  =  7 


By  eqs.  (3)  and  (4), 


Integrating, 

C£2  -  r2)  log,  x  =  2r2  log,  g  +  c, 

c  being  constant  of  integration. 
When  x  =  D,  z  —  r, 

/.  (^2  -  r2)  log,  -^  =  2r3  log,  |, 
or 


WORK.  167 

By  eqs.  (2)  and  (5), 


which  is  the  equation  to  the  profile  PL  or  QK. 

Note.  —  If  R-  —  3r2,  eq.  (6)  represents  a  straight  line. 
"  R*  =  2r\       4<  "  "  parabola. 

II.  Work.  —  Work  must  be  done  to  overcome  a  resistance. 
Thus  bodies,  or  systems  of  bodies,  which  have  their  parts  suit- 
ably arranged  to  overcome  resistances  are  capable  of  doing- 
work  and  are  said  to  possess  energy.  This  energy  is  termed 
kinetic  or  potential  according  as  it  is  due  to  motion  or  to  posi- 
tion. A  pile-driver  falling  from  a  height  upon  the  head  of  a 
pile  drives  the  pile  into  the  soil,  doing  work  in  virtue  of  its 
motion.  Examples  of  potential  energy,  or  energy  at  rest,  are 
afforded  by  a  bent  spring,  which  does  work  when  allowed  to 
resume  its  natural  form  ;  a  raised  weight,  which  can  do  work  by 
falling  to  a  lower  level  ;  gunpowder  and  dynamite,  which  do 
work  by  exploding  ;  a  Leydenjar  charged  with  electricity,  which 
does  work  by  being  discharged  ;  coal,  storage  batteries,  a  head 
of  water,  etc.  It  is  also  evident  that  this  potential  energy 
must  be  converted  into  kinetic  energy  before  work  can  be 
done.  A  familiar  example  of  this  transformation  may  be  seen 
in  the  action  of  a  common*  pendulum.  At  the  end  of  the 
swing  it  is  at  rest  for  a  moment  and  all  its  energy  is  potential. 
When,  under  the  action  of  gravity,  it  has  reached  the  lowest 
point,  it  can  do  no  more  work  in  virtue  of  its  position.  It  has 
acquired,  however,  a  certain  velocity,  and  in  virtue  of  this 
velocity  it  does  work  which  enables  it  to  rise  on  the  other  side 
of  the  swing.  At  intermediate  points  its  energy  is  partly 
kinetic  and  partly  potential. 

A  measure  of  energy,  or  of  the  capacity  for  doing  work,  is 
the  work  done. 

The  energy  is  exactly  equivalent  to  the  actual  work  done 
in  the  following  cases: 

(a)  If  the  effort  exerted  and  the  resistance  have  a  common 
point  of  application. 


1 68  THEORY  OF  STRUCTURES. 

(b)  If  the  points  of  application  are  different  but  are  rigidly 
connected. 

(c)  If  the  energy  is  transmitted  from  member  to  member, 
provided  the  members  do  not  change  form  under  stress,  and 
that  no  energy  is  absorbed  by  frictional  resistance  or  restraint 
at  the  connections. 

Generally  speaking,  work  is  of  two  kinds,  viz.,  internal  work, 
or  work  done  against  the  mutual  forces  exerted  between  the 
molecules  of  a  body  or  system  of  bodies,  and  external  work,  or 
work  done  by  or  against  the  external  forces  to  which  the  body 
or  bodies  are  subjected.  In  cases  (a),  (b),  (c)  above,  the  inter- 
nal work  is  necessarily  nil. 

As  a  matter  of  fact,  every  body  yields  to  some  extent  under 
stress,  and  work  must  be  done  to  produce  the  deformation. 
Frictional  resistances  tend  to  oppose  the  relative  motions  of 
members  and  must  also  absorb  energy.  If,  however,  the  work 
of  deformation  and  the  work  absorbed  by  frictional  resistance 
are  included  in  the  term  work  done,  the  relation  still  holds  that 

Energy  =  work  done. 

A  measure  of  work  done  is  the  product  of  the  resistance  by 
the  distance  through  which  it  is  overcome.  When  a  man 
raises  a  weight  of  one  pound  one  foot  against  the  action  of 
gravity  he  does  a  certain  amount  of  work.  To  raise  it  two  feet 
he  must  do  twice  as  much  work,  arid  ten  times  as  much  to  raise 
it  ten  feet.  The  amount  of  work  must  therefore  be  propor- 
tional to  the  number  of  feet  through  which  the  weight  is 
raised.  Again,  to  raise  two  pounds  one  foot  requires  twice  a.* 
much  work  as  to  raise  one  pound  through  the  same  distance ; 
while  five  times  as  much  work  would  be  required  to  raise  five 
pounds,  and  ten  times  as  much  to  raise  ten  pounds.  Thus  the 
amount  of  work  must  also  be  proportional  to  the  weight  raised. 
Hence  a  measure  of  the  work  done  is  the  product  of  the 
number  of  pounds  by  the  number  of  feet  through  which  they 
are  raised,  the  resulting  number  being  designated  foot-pounds. 
Any  other  units,  e.g.,  a  pound  and  an  inch,  a  ton  and  an 
inch,  a  kilogramme  and  a  metre,  etc.,  may  be  chosen,  and  the 
work  done  represented  in  inch-pounds,  inch-tons,  kilogram- 


OBLIQUE   RESISTANCE. 


169 


metres,  etc.  This  standard  of  measurement  is  applicable  to  all 
classes  of  machinery,  since  every  machine  might  be  worked  by 
means  of  a  pulley  driven  by  a  falling  weight. 

12.  Oblique  Resistance.  —  Let  a  body  move  against  a 
resistance  R  inclined  at  an  angle  0  to  the  direction  of  motion 
(Fig.  189).  No  work  is  done  against  the 
normal  component  R  sin  0,  as  there  is 
no  movement  of  the  point  of  applica- 
tion at  right  angles  to  the  direction  of 
motion.  This  component  is,  there- 
fore. merely  a  pressure.  The  work 
done  against  the  tangential  component 
R  .  cos  6  between  two  consecutive 
points  M  and  N  of  the  path  of  the  body  is  R  cos  6  .  MN. 
Hence  the  total  work  done  between  any  two  points  A  and  B  of 
the  path 

=  2(R  cos  0  .  MN)  =      SR  cos  6ds, 


FIG.  i 


s  being  the  length  of  AB. 

If  AB  is  a  straight  line  (Fig.  190),  and  if  R  is  constant  in 
direction  and  magnitude, 

the  total  work  =  R  cos  B  .AB  —  R.AC, 

AC  being  the  projection  of  the  displacement  upon  the  line  of 
action  of  the  resistance.     Let  the  path   be  the  arc  of  a  circle 


Rsintf 


FIG.  190.  FIG.  191. 

(Fig.  191)  subtending  an  angle  a  at  the  centre.     If  R  and  6  re- 
main constant,  the  work  done  from  A  to  B 


=  R  cos  0  arc  AB  =  R  cos  0  .  OA  .  a  =  R  .  OM  cos  6  .  a 


I7O  THEORY  OF  STRUCTURES. 

p  being  the  perpendicular  from  0  upon  the  direction  of  R,  and 
M  =  Rp  being  the  moment  of  resistance  to  rotation. 

If  there  are  more  resistances  than  one,  they  may  be  treated 
separately  and  their  several  effects  superposed.  In  such  case, 
Mwi\\  be  the  total  moment  of  resistance  and  will  be  equal  to 
the  algebraic  sum  of  the  separate  moments. 

The  normal  component  R  sin  8  produces  a  pressure. 

13.  Graphical  Method. — Let  a  body  describe  a  path  AB 

B  (Fig.  I92)against  a  variable  resistance  of 
such  a  character  that  its  magnitude  in 
the  direction  of  motion  may  be  repre- 
sented at  any  point  M  by  an  ordinate 

'D  MNto  the  curve  CD.  Let  the  path 
AB  be  subdivided  into  a  number  of 
parts,  each  part  MP  being  so  small 
that  the  resistance  from  M  to  P  may 
be  considered  uniform.  The  mean 

value  of  this  resistance  —  -  — ,  and  the  work  done  in 

MN-\-PQ 
overcoming  it  = —  .  MP  =  the  area  MNQP  in  the 

limit.  Hence  the  total  work  done  from  A  to  B  =  the  area 
bounded  by  the  curves  AB,  CD  and  the  ordinates  AC,  BD. 

14.  Kinetic  Energy. — The  velocity  v  acquired  by  a  body 
of  weight  w  and  mass  m  in  falling  freely  from  rest  through  the 
vertical  distance  h  is 

v  =   \/2gh ; 

w  v          v* 

/.  wh  = =  m— . 

g2  2 

Thus  an  amount  of  work  wh  is  done,  and  the  body  possesses 

v* 
the  kinetic  energy  m— . 

Again,  let  v'  be  the  velocity  of  the  body  after  falling 
through  a  further  distance  x,  measured  vertically.  Then 


w(/i  -j-  x)  — 


KINETIC  ENERGY. 
and 


Thus  the  work  done  in  falling  through  the  vertical  distance  x 
is  wx,  and  is  equal  to  the  corresponding  change  of  kinetic 
energy. 

15.  EXAMPLE  I.  Let  it  be  required  to  determine  the  work 
done  in  stretching  or  compressing  a  bar  of  length  L  and  sec- 
tional area  A  by  an  amount  /. 

Suppose  that  the  force  applied  to  the  bar  gradually  in- 
creases from  o  until  it  attains  the  value  P\  its  mean  value  is 

P  P 

—  ,  and  the  work  done  is  therefore  —  /. 
2'  2 

But  P  =  EA-   ;  E  being  the  coefficient  of  elasticity. 


E    /"       i  IP\ 

/.  the  work  done  =  —A-?  =  •55-1-3"] 
2     L      E\Ai 


AL_ 

2 


/a 

This  formula  is  only  true  for  small  values  of  the  ratio  j. 

In  the  case  of  a  compressive  force  it  is  assumed  that  the  bar 
does  not  bend. 

P 
A  suddenly  applied  force,  —  ,  will  do  as  much   work  as  a 

steady  force  which  increases  uniformly  from  o  to  P,  and  hence 
it  follows  that  a  bar  requires  twice  the  strength  to  resist  with 
safety  the  sudden  application  of  a  given  load  than  is  necessary 
when  the  same  load  is  gradually  applied. 

If /is  the  proof  stress  or  elastic  limit  per  unit  of  sectional 

area,  7^  is  the  corresponding  proof  strain,  and  the  work  done  in 

£L 

producing  the  latter  is  called  the  resilience  of  the  bar.     Accord- 

f*  AL     f* 
ing  to  the  above,  its  value  is  ~  —  ;  ~-  is  called  the  Modulus 

of  Resilience. 


172  THEORY  OF  STRUCTURES. 

Ex.  2.  A  wrought-iron  tie-rod,  30  ft.  in  length  and  4  sq. 
in.  in  sectional  area,  is  subjected  to  a  longitudinal  pull  of 
40,000  Ibs.  Determine  the  unit  stress,  the  strain,  and  the  elon- 
gation, the  coefficient  of  elasticity  being  30,000,000  Ibs. 

40000 

The  unit  stress  is =  10,000  Ibs.  per  sq.  in. 

4 

Also,  from  the  elastic  law,  10000  —  30000000  X  strain. 

.-.  the  strain  =  ~ 
and  the  elongation  =  ~  —  ^  ft. 

Ex.  3.  A  steel  rod  is  15  ft.  long  and  2\  sq.  in.  in  sec- 
tional area.  The  proof  strain  of  the  steel  is  ^,  and  its  coeffi- 
cient of  elasticity  is  36,000,000  Ibs.  Find  the  greatest  weight 
that  can  be  safely  allowed  to  fall  upon  the  end  of  the  rod  from 
a  height  of  27  ft?. 

The  proof  stress  =  Ex  proof  strain  =  36,000  Ibs.  persq.  in. 

The  compression  of  the  rod  under  the  proof-stress  is 

Jl  _JL  ft 

1000  —  200   *•*•• 

The  resilience  of  the  rod 

_f*AL  _  (36000)'  2J-  X  15  X  12 

~  E     2  36000000  2 

=  8100  inch-lbs.  =  675  ft.-lbs. 

Again,  let  P^be  the  required  weight  in  pounds. 

The  total  distance  through  which  it  falls  —  27  ft.  -f-  com- 
pression =  (27  -f-  TO)  ^eet>  anc*  the  corresponding  work  is 
W(2J  +  4)  ft.-lbs.  This  must  of  course  be  exactly  equivalent 

to  the  resilience  of  the  rod,  and 

* 

•••  W(27  +  £)  =  675, 
and  W  —  24.9  Ibs. 


KIXETIC  ENERGY.  173 

Tht  resilience  of  the  rod  may  also  be  at  once  found  from 
the  fact  that  it  is  the  product  of  one  half  of  the  total  stress  by 
the  compression,  i.e.,  -J-  .  2|-  .  36000  X  ^  =  675  ft.-lbs. 

Ex.  4.  Let  w1  ,  w2  ,  w3  ,  .  .  .  wn  be  the  weights  of  a  system 
of  particles  rigidly  connected  together  and  at  distances  ^  ,  jra  , 
JTS  ,  .  .  .  xn  ,  respectively,  from  a  given  axis.  Let  the  system 
revolve  around  the  axis  with  a  uniform  angular  velocity  A. 

The  kinetic  energies  of  the  several  particles  are 

w  xA*       w  x*A*  w   x*A* 


and  therefore  the  total  kinetic  energy  of  the  system 


r, 

A*  ( 
=  —  -  w.^if 

m^  ,  m^  ,  .  .  .  mn  being  the  masses  of  the  particles. 

The  sum  between  the  brackets  is  called  the  moment  of  in- 
ertia of  the  system  of  particles  about  the  axis  and  is  usually 
denoted  by  /. 

A*  I 
.'.  the  total  kinetic  energy—  -—  . 

Again,  it  appears  from  the  definition  that  every  moment  of 
inertia  is  the  product  of  a  mass  and  the  square  of  a  length. 
This  length  is  called  the  radius  of  gyration  and  is  usually 
designated  by  the  symbol  k. 

If  M  be  the  total  mass  of  the  system,  and  W  the  total 
weight, 

W 


M£)2       W  (Ak? 
and  the    total    kinetic   energy  =  M  -  -  =  —  -  -  —,  the  re- 


174  THEORY  OF  STRUCTURES. 

suit  being  the  same  as  if  the  particles  were  collected  in  a  ring 
of  radius  k,  sometimes  called  the  equivalent  ring  or  fly-wheel. 

Let  Ig  be  the  moment  of  inertia  of  the  system  with  respect 
to  a  parallel  axis  through  the  centre  of  gravity,  and  let  h  be  the 
distance  between  the  two  axes.  Then 

IK  =  >«,(/<  -  *,)'  +  mjji  -  O2  +  .  .  .  +  mn(h  -  xnf 


Since  the  new  axis  passes  through  the  centre  of  gravity, 

2mx  =  Mh. 
Also,  2(m)  =  M    and     2(m^)  =  7; 


.'.  I=Ig+Mh\ 

So,  if  /'  is  the  moment  of  inertia  about  another  parallel  axis 
at  the  distance  h'  from  the  centre  of  gravity, 

I'  =Ie+Mh'\ 
.-.  I-  Mh*  =  I'  —  Mh'\ 


Hence,  if  the  positions  of  two  parallel  axes  relatively  to  the 
centre  of  gravity  are  known,  and  if  the  moment  of  inertia  about 
one  is  given,  the  moment  of  inertia  about  the  other  can  be 
obtained  by  means  of  the  last  formula. 

Note.  —  Nothing  has  been  said  as  to  the  number  of  the  par- 
ticles. They  may  be  infinite  in  number  and  infinitely  near 
each  other,  forming  in  fact  a  solid  body.  The  summation 
2(mx*)  is  then  best  effected  by  integration. 

16.  Values  of  w. 

1.  For  a  rectangular  plate  of  depth  d  with  re- 

spect   to   an  axis  through    the  centre  ,2 

perpendicular  to  the  side  d.  ..........   k*  =  —  . 

2.  For  a  circular  plate  of  radius  r  with  re-  a 

spect  to  a  diameter  ..................   k*  =  -. 

4 


VALUES   OF  k\  175 


3.  For  an  annulus  of  external  radius  rl  and 
internal  radius  r^  with  respect  to  a 
diameter  .......................... 


.  —  If  r1  —  r^  =  t,  and  the  breadth 
/  of  the  annulus  is  small  as  compared 
with  the  radius  rl  ,  then 


approx., 

4  ^ 

and  the  area 

approx. 


4.  For  the  plates  in  (2)  and  (3)  with  respect 

to  an  axis  through  the  centre  perpen- 
dicular to  the  plates,  the  numerators 
remain  the  same  but  the  denominator 
is  in  each  case  2. 

5.  For  a  sphere  of  radius  r  with  respect  to  a 

diameter k*  =  —r*. 

6.  For  a  solid  cylinder  of  radius  r  with  re-  3 

spect  to  its  axis k*  =  — . 

2 

7.  For  an  elliptic  plate   of  which  the  major 

and  minor  axes  are  2b  and  2d  respec- 
tively : 

With  respect  to  the  major  axis k*  =  — . 

4 

ia 

With  respect  to  the  minor  axis J?  =  — . 

4 

8.  For  a  triangular  plate  of  height  h  with  re- 

spect  to    an    axis  coinciding  with  the  ,a 

base . .' k*  =  —  -. 

6 


THEORY  OF  STRUCTURES. 


17.  Momentum—  Impulse.  —  A  moving  body  of  weight  w 
and  mass  m  acted  upon  in  the  direction  of  motion  for  a  time  / 
by  a  force  F  will  acquire  a  velocity  v  which  is  directly  propor- 
tional to  F  and  to  /,  and  inversely  proportional  to  w.  Hence 


Ft 

v  =  n  — 


n  being  some  coefficient. 

If  F  =  w,  the  velocity  generated  in  one  second 

•'.  g  =  n, 
and 

Ft       Ft 

•''"  =  ^=-m> 
or 

mv  =  Ft. 

This  is  the  analytical  statement  of  Newton's  Second  Law 
of  Motion,  which  has  been  expressed  by  Clerk  Maxwell  in  the 
following  form  :  "  The  change  of  momentum  (i.e.,  the  product  of 
the  mass  and  velocity)  is  numerically  equal  to  the  impulse  (i.e., 
the  product  of  the  force  and  the  time  during  which  it  acts) 
which  produces  it,  and  is  in  the  same  direction" 

Again,  let  /  be  the  perpendicular  from  a  fixed  axis  O  upon 
the  direction  of  motion  of  the  body,  and  let  r  be  the  radius  OP 
to  the  body.  Then 

mvp  =  Ftp  =  Fpt  =  Mt, 

where  M  =  Fp  ;  or  the  change  of  the  moment  of  momentum,  i.e., 
of  the  angular  momentum,  is  equal  to  the  moment  of  impulse. 

The  above  results  are  also  true  for  two  or  more  bodies  or 
systems  of  bodies  severally  acted  upon  by  extraneous  forces, 
and  the  equations  may  be  written 


In  words,  the  total  change  of  momentum  in  any  assigned  direction 
is  equal  to  the  algebraic  sum  of  the  impulses  in  the  same  direction, 


MOMENTUM— IMPULSE.  1/7 

and  the  total  change  of  angular  momentum  is  equal  to  the  alge- 
braic sum  of  the  moments  of  the  impulses. 

Hence  it  follows  that  if  two  or  more  bodies  or  systems  of 
bodies  mutually  attract  or  repel  each  other,  and  if  there  are  no 

extraneous  forces,  the  total  momentum  in      ^  w 

*^-»  fc * 

any   assigned   direction    is   constant   (the          ~"7^^        \ 

principle  of   the   conservation   of   linear  /  *s***^p 

momentum),  and  the  angular  momentum         j  /'' 

about  a  given  axis  is  constant  (the  prin-         / :        r/' 
ciple    of    the    conservation    of    angular       /      /' 
momentum).  ,//'  f2  /  a 

Suppose  that  the  velocity  of  the  body  3^-- -^ y 

of  weight  w  and  mass  ;//  changes  from  X^  / 

vl  to  vz  in  the  time  /  under  the  action  of  2^Xv// 

a  couple  of  moment  M,  and  let/,,/,  be  FlG-  ^s- 

the  corresponding  values  of/,  and  rlf  ra  those  of  r,  Fig.  193. 


or  if  wlt  w^  are  the  components  of  vl ,  v^  in  directions  perpen- 
dicular to  rlt  7*a,  respectively, 

m(w1rl  —  wj-^  —  Mt. 

For  example,  a  weight  W  of  water  passing  through  a  turbine 
of  external  radius  r^  and  internal  radius  r^  has  its  angular  mo- 

W  W 

mentum    changed    from   — wlrl    to   — ^ara>    wi->    w*     being 

o  o 

the  tangential    components   of   the  velocity  with   which    the 
water  enters  and  leaves  the  wheel.    The  water,  therefore,  exerts 

W 
upon   the  wheel  a  couple  of  moment  —  (lwlrl  —  w2^2),  and    if 

o 

the  wheel  rotates  with  an  angular  velocity  A,  the  work  done 
upon  the  wheel  by  the  water 

w  w 

=  ~A(wj-i  —  wyj  =  —  (ze/.a,  —  w,u,), 

<b  O 

Jil  and  u^  being  the  circumferential  velocities  corresponding  to 
rl  and  ra,  respectively. 


THEORY  OF  STRUCTURES. 

18.  Useful  Work— Waste  Work. — Let  a  body  of  mass  m 
.and  weight  w  pass  over  the  distance  s  under  the  action  of  a 
force  F  acting  in  the  direction  of  motion  for  a  time  /,  and  let 
the  velocity  of  the  body  change  from  vl  to  v9.  Assume  /  to 
.be  so  small  that,  for  the  interval  in  question,  the  velocity  may 

i)    I   ij 
>be  regarded  as  constant  and  of  the  average  value  -L— • — - ; 


But  Ft  =  (mvt  —  mVt). 


m     , 
2^  * 


or 


Fs  =     (»,'  -  »,•). 

Thus  Fs,  the  work  done,  is  equal  to  the  change  of  kinetic 
energy  in  the  given  interval. 

If  the  body  is  a  material  particle  of  a  connected  system,  a 
similar  relation  holds  for  every  other  particle  of  the  system,  and 
the  total  work  done  =  %(2mv*  —  2mv?). 

A  part  of  this  work  may  be  expended  in  doing  what  is 
called  effective  work,  i.e.,  in  overcoming  (i)  an  external  resist- 
ance, or  in  doing  useful  work,  and  (2)  frictional  resistance,  or  in 
doing  wasted  work. 

Denoting  the  total  effective  work  by  Te  and  the  total  motive 
work  by  Tm  ,  the  last  equation  may  be  written 


and  the  difference  between  the  total  motive  work  and  the  total 
effective  work  is  equal  to  the  total  change  of  kinetic  energy. 

In  the  case  of  a  machine  working  at  a  normal  speed  the 
velocities  of  the  different  parts  are  periodic,  being  the  same  at 
the  beginning  and  end  of  any  period  or  number  of  periods. 
For  any  such  interval,  therefore,  vt  =  vy,  and  /.  Tm  —  Te,  so 


GENERAL    CASE.  1  79 

that  there  is  an  equality  between  the  motive  work  and  the 
effective  work. 

19.  General  Case.  —  Let  xl  ,  ylt  z^  be  the  co-ordinates  of 
the  C.  of  G.  of  a  moving  body  of  mass  M  with  respect  to  three 
rectangular  axes  at  any  given  instant. 

Let  xt,  yt,  £2  be  the  co-ordinates  of  the  same  point  after  a 
unit  of  time. 

Let  x,  ,  yl  ,  z,  be  the  co-ordinates  of  any  particle  of  mass  m 
at  the  given  instant. 

Let  x^,y^,  £2  be  the  co-ordinates  of  the  same  particle  after 
a  unit  of  time. 


,  -  y,}  =  2m(y*  -  y, 


or 

M  u  =  2mu,     Mv  —  2mv,     Mw  = 

u,  v,  w  being  the  component  velocities  of  the  C.  of  G.  at  the 
given  instant  with  respect  to  the  three  axes,  and  u,  v,  w  the 
component  velocities  of  the  particle  m  at  the  same  instant. 
From  these  last  equations, 

Mu  =  ^muu,     Mv  —  2mvv,     Mw   = 


.'.  M(u  -}-v'-\-w)  =  2m(uu  -\-  vv 
which  may  be  written  in  the  form 


., — 2       .      — 2 


M(u  +  <v  -\-  w)  +  2m\(u  -  uj  +  (v  -  v?  +  (w  - 

=  2m(u*  +  v*  +  w2), 


or 


ISO  THEORY  OF  STRUCTURES, 

U  being  the  resultant  velocity  of  the  C.  of  G. ;  v,  that  of  the 
particle  ;  and  F,  that  of  the  particle  relatively  to  the  C.  of  G. 
The  last  equation  may  be  written 


MIT       ^mV*  _2mv> 

O  I  1  /> 


Thus  the  energy  of  the  total  mass  collected  at  the  centre 
of  gravity,  together  with  the  energy  relatively  to  the  centre  of 
gravity,  is  equal  to  the  total  energy  of  motion. 

If  the  body  revolves  around  an  axis  through  its  C.  of  G. 
with  an  angular  velocity  A,  the  second  term  of  the  last  equa- 
tion becomes 


r  being  the  distance  of  the  particle  m  from  the  axis,  and  /  the 
moment  of  inertia  of  the  body  with  respect  to  the  axis. 

20.  EXAMPLE  I.  The  charge  of  powder  for  a  27-ton  breech- 
loader with  a  Q-ton  carriage  is  300  Ibs.  ;  the  weight  of  the  pro- 
jectile is  500  Ibs.,  its  diam.  is  10  in.,  and  its  radius  of  gyration 
3.535  in.  ;  the  muzzle  velocity  is  2020  ft.  per  sec.  ;  the  velocity 
of  recoil,  i6J  ft.  per  sec.  ;  the  gun  is  rifled  so  that  the  projectile 
makes  one  turn  in  40  calibres. 

Total  energy  of  explosion  =  energy  of  shot  -|-  energy  of  recoil  : 
Energy  of  shot  =  energy  of  translation  +  energy  of   rotation 

_  5Q      (2020)'       500    j.    /V.-j-f     2020  y  /3-535\a 

32.2''       2  t"32.2'2*\    T82-      "4Q.|jy     \     12    I 

=  31680124.2  +97758.6 
=  31777882.8  ft.-lbs.  ; 


En.ro  of  recoil  *  .  .  330652,  ft,.b, 


CENTRIFUGAL   FORCE.  l8l 

Hence,  if  C  be  the  energy  of  I  Ib.  of  powder, 

C .  300  =  31777882.8  +  330652.1 

=  32108534.9  ft.-lbs., 
and  hence 

C  —  107028.45  ft.-lbs  =  47.7  ft.-tons. 

Ex.  2.  Let  Wbe  the  weight  of  a  fly-wheel  in  Ibs,,  and  let 
its  max.  and  min.  angular  velocities  be  AltA9,  respectively. 
The  motion  being  one  of  rotation  only,  the  energy  stored  up 
when  the  velocity  rises  from  A^  to  Alt  or  given  out  when  it 
falls  from  Al  to  A^  is 

I  W  W 

-  (A:  -  A:)  =  -k\A?  -  A:)  =  —  w  -  *.•), 

vl ,  v^  being  the  linear  velocities  corresponding  to  Alt  A2,  and 
k  being  taken  equal  to  the  mean  radius  of  the  wheel. 

It  is  usual  to  specify  that  the  variation  of  velocity  is  not 
to  exceed  a  certain  fractional  part  of  the  mean  velocity. 

Let  V  be  the  mean  velocity,  and  —  the  fraction.     Then 

V 
Vi  —  v*  =  —'>     also    vl  +  vt 


W  Fa 

Hence  the  work  stored  or  given  out  = . 

*  g    P 


21.  Centrifugal   Force. — A  body  constrained  to  move  in 
a  plane  curve  exerts  upon  the  body  which  constrains  it,  a  force 


1  82  THEORY   OF  STRUCTURES, 

called  centrifugal  force,  which  is  equal  and  opposite  to  the  de- 
viating (or  centripetal}  force  exerted  by 
the  constraining  body  upon  the  revolving 
body. 

Let  a  particle  of  mass  m  move  from  a 
\  point  P  to  a  consecutive  point  Q  (Fig.  194) 

\  of  its  path  during  an  interval  of  time  / 

\o  under  the  action  of  a  normal  deviating 

FIG.  i94.  force. 

Let  the  normals  at  P  and  Q  meet  in  O  ;  PQ  may  be  con- 
sidered as  the  indefinitely  small  arc  of  a  circle  with  its  centre 
at  O. 

If  there  were  no  constraining  force,  the  body  would  move 
along  the  tangent  at  P  to  a  point  T  such  that  PT  =  vt,  v 
being1  the  linear  velocity  at  P. 

Under  the  deviating  force  the  body  is  pulled  towards  O 
through  a  distance  PN  =  %ff,  f  being  the  normal  accelera- 
tion, and  QN  being  drawn  perpendicular  to  OP. 
Also,  in  the  limit,     PQ  =  PT  =  QN  =  vt. 

But 


being  the  radius  OP;  and  hence 


A  being  the  angular  velocity. 

Hence  the  deviating  force  of  the  mass  m 


=  mf  =  m  -=  =  mA*ry 

and  is  equal  and  opposite  to  the  centrifugal  force. 

Again,  if  a  solid  body  of  mass  M  revolve  with  an  angular 
velocity  A  about  an  axis  passing  through  its  C.  of  G.,  the  total 


CENTRIFUGAL   FORCE. 


183 


centrifugal  force  will  be  nil,  provided  the  axis  of  rotation  is  an 
axis  of  symmetry,  or  is  one  of  the  principal  axes  of  inertia  at 
the  C.  of  G. 

If  the  axis  of  rotation  is  parallel  to  one  of  these  axes,  but 
at  a  distance  R  from  the  C.  of  G., 


the  centrifu- 

,  , 
gal  force 


W 

— 

g 


r  being  the  distance  of  a  particle  of  mass  m  from  the  axis,  and 
f'Fthe  weight  of  the  body.  Thus  the  centrifugal  force  is  the 
same  as  if  the  whole  mass  were  concentrated  at  the  C.  of  G. 

If  the  axis  of  rotation  is  inclined  at  an  angle  0  to  the  prin- 
cipal   axis,    the   body  will    be    con- 
stantly subjected    to    the   action    of 
a   couple   of   moment    2E  tan  6,  E 
being  the  actual  energy  of  the  body. 

EXAMPLE.  —  A  ring  of  radius  r 
rotates  with  angular  velocity  A  about 
its  centre  O.  Let  p  be  the  weight  of 
the  ring  per  unit  of  length  of  periph- 
ery. Consider  any  half-ring  AFB. 
The  centrifugal  force  of  any  element 


FlG-  I9S> 


The  component  of  this  force  parallel  to  AB,  is  balanced  by 
an  equal  and  opposite  force  at  C"  ,  the  angle  C"  OB  being  — 
the  angle  CO  A.  Thus  the  total  centrifugal  force  parallel  to 
A  OB  is  nil. 

The  component  of  the  force  at  C,  perpendicular  to  AB, 


g 


sn 


COD=P—-A"r  cos  C'CE 


_fCC        C^_      A'r 
—Arcc,-p--DD. 


184  THEORY  OF   STRUCTURES. 

Hence,  the  total  centrifugal  force  perpendicular  to  AB 

=  ?—2(DDf)  =  2^4  V. 
g  g 

If  T  is  the  force  developed  in  the  material  at  each  of  the 
points  A  and  B, 


g 

since  the  direction  of  T  is  evidently  perpendicular  to  AB. 

~~g          '  F  ' 

v  being  the  circumferential  velocity. 

Let  f  be  the  intensity  of  stress  at  A  and  B,  and  w  the 
specific  weight  of  the  material. 

Assuming  that  T  is  distributed  uniformly  over  the  sectional 
areas  at  A  and  B, 


Thus,  the  stress  is  independent  of  the  radius  .for  a  given 
value  of  v,  and  the  result  is  applicable  to  every  point  of  a  flex- 
ible element,  whatever  may  be  the  form  of  the  surfaces  over 
which  it  is  stretched. 

22.  Impact. — When  a  body  strikes  a  structure,  or  member 
of  a  structure,  the  energy  of  the  blow  is  expended  in 

(1)  overcoming  the  resistance  to  motion  of  the  body  struck  ; 

(2)  deforming  the  body  struck  ; 

(3)  the  kinetic  energy  of  either  or  of  both  of  the  bodies 
after  impact,  if  the  motion  is  sensible  ; 

(4)  deforming  the  striking  body ; 

(5)  producing  vibrations. 


IMPACT.  185 

Generally  speaking,  the  energy  represented  by  (5)  is  very 
small  and  may  be  disregarded.  Also,  if  the  striking  body  is 
very  hard,  the  energy  (4),  absorbed  in  its  deformation,  is  inap- 
preciable and  may  be  neglected. 

First,  let  a  body  of  weight  P  fall  through  a  vertical  dis- 
tance h  and  strike  a  second  body,  the  point  of  application 
moving  in  the  direction  of  the  blow  through  a  distance  x 
against  a  mean  resistance  R.  Then 

P(k  -f-  x)  =  work  done  =  R'x. 

Let  Fbe  the  velocity  of  the  striking  body  at  the  moment 
of  impact.  Then 


P  V 
energy  of  blow  =  -  -  —  =  R'x  =  P(1i  -\-  x). 


The  actual  resistance  is  directly  proportional  to  the  dis- 
tance through  which  the  point  of  application  moves,  so  long  as 
tJie  limit  of  elasticity  is  not  exceeded.  Its  initial  value  is  nil, 

r> 

and  if  R  is  its  max.  value,  the  mean  value  is  R'  =  — . 


P  V       Rx 
.'.  -    -  =  -  -  =  P(h  +  x). 

g    2  2  ' 


If  h  =  o,  R  =  2P,  or  the  sudden  application  of  a  load  P 
from  rest,  produces  a  pressure  equal  to  twice  the  load,  pro- 
vided the  limit  of  elasticity  is  not  exceeded. 

EXAMPLE.  A  i-oz.  bullet  moving  with  a  velocity  of  800  ft. 
per  sec.  strikes  a  target  and  is  stopped  dead  in  the  space  of 
,ft.  inch  (g  =  32),  Then 


1 86  THEORY   OF  STRUCTURES. 

.-.  R ',  the  mean  resistance  overcome  by  the  bullet,  =  5000  Ibs. 
The  time  in  which  the  bullet  is  brought  to  rest 


momentum      y1-^ .  •£% .  800         I 

force  5000         ~  3200  SeC* 

Next,  let  a  body  of  weight  W^  moving  in  a  given  direction 
with  a  velocity  z\  strike  a  body  of  weight  W^  moving  in  the 
same  direction  with  a  velocity  v^ .  After  impact  let  the  bodies 
continue  to  move  in  the  same  direction  with  a  common  ve- 
locity v. 

W,  W, 

—  P,  +  —  v+  =  momentum  before  impact 

=  momentum  after  impact 


or 


,        ,       ^ 
Energy  before  impact  =  —  —  +  —  — . 


after        "        =1— ^ -)— . 

g 


Energy  lost  by  impact 

% 


IMPACT.  187 

If  either  of  the  bodies  is  subjected  to  any  constraint,  energy 
must  be  expended  to  overcome  such  constraint,  and  the  loss  of 
energy  by  impact  will  be  less. 

EXAMPLE  i.  Let  a  weight  of  W^  tons  fall  h  ft.  upon  the 
head  of  a  pile  weighing  W^  tons  and  drive  it  a  ft.  into  the 
ground  against  a  mean  resistance  of  R  tons,  the  head  of  the 
pile  being  crushed  for  an  appreciable  length  x  ft. 

Let  v  be  the  velocity  of  the  weight  when  it  strikes  the  pile  ; 
n   p  n     «     mean  force  of  the  blow; 
"    y   "     "     distance  through  which  pile  moves  during  ac- 

tion of  blow  ; 

"    /   "     "     duration  of  the  blow  in  seconds  ; 
"   V  "     "     common  velocity  of  the  pile  and  weight  during 

action  of  blow  ; 
"  z    "     "     distance  through  which  pile  moves  after  the 

blow. 

Px  -\-  Ry  =  work  done  in  crushing  the  pile  -f-  work 
done  in  overcoming  ground-resistance 
in  time  t  =  energy  dissipated  by  blow 


Also,  considering  the  change  of  momentum  first  of  weight 
and  then  of  pile, 


Again, 

W    I    W  I7* 
Rz  =  work  done  after  blow  =  — — 1- — -  — .  (*\ 

g         2 

Finally,  y  +  2  =  a,   . (4) 

and 

(5) 


1  88  THEORY  OF  STRUCTURES. 

Thus,  if  Wlt  Wif  h,  a,  and  x  are  known,  eqs.  (i)  to  (5)  will 
give  P,  /,  R,  y,  V,  and  z. 

Ex.  2.  Let  a  hammer  weighing  Wl  Ibs.  moving  with  a  ve- 
locity of  v  ft.  per  sec.,  strike  a  nail  weighing  W^  Ibs.  and  drive 
it  x  ft.  into  a  piece  of  timber,  of  weight  W3J  against  a  mean 
resistance  of  R  Ibs. 

First,  assume  the  timber  to  be  fixed  in  position. 

Let  F,  be  the  common  velocity  acquired  by  the  hammer 
and  nail. 


\  ~f"  ^a)  —  L  —  energy  expended  in  overcoming  R 


(i) 


W  W  4-  W 

But         —  -v  —  change  of  momentum  =  —  —-  -  ?F1.       .     (2) 

o  o 


3 


x         Wv 
and  the  time  of  the  penetration  =  JT>~  =  -jr-  sec.      ...     (4) 

Second,  let  the  timber  be  free  to  move,  and  let  F2  be  the 
common  velocity  acquired  by  the  hammer,  nail,  anld  timber. 

F2 
(W\~\~  W^)  —  =  energy  expended  in  overcoming  R  plus 

the  energy  expended  in  producing  the 
velocity  F2 


.  ...     (5) 


W          w  _i_  w  W  4-  W  -4-  W 

But  ^^^TJ^F^-^^J    --^   ^F2.     .     .     (6) 

g  g  g  V  ; 


ON    THE  EXTENSION  OF  A    PRISMATIC  BAR.  189 

Hence,  substituting  these  values  of  F,  and  F3  in  eq.  (5), 


J?v    .  f>7\ 

—  =  Kx\     .     .     .     (7) 


also,  the  time  of  the  penetration 


V_  .    . 

* 


and  the  distance  through  which  the  timber  moves 

W?W*  jLft 

U 


23.  On  the  Extension  of  a  Prismatic  Bar.  —  The  ele- 
mentary law  of  extension  is  sometimes  enunciated  as  follows: 

A  prismatic  bar  of  length  L  and  sectional  area  A  is 
stretched,  and  its  length  is  L  -\-  x  when  the  force  of  extension 
isP-  if  dP  is  the  increment  of  force  corresponding  to  an  in- 
crement dx  of  length, 

dP=EA     d*. 

L  +  x 

Hence,  the  force  producing  an  extension  /  is  equal  to 
j  =  EA  log.(i  +~)=^,  s 


suppose. 

But 

/         A       /       i/Aa      i//\3  / 

aPProx' 


/         A       /       i/Aa      i//\3  / 

+  IJ  =  L  ~  2  (ll  +  jlzJ  -  •  •  •  =  Z  ' 


I  QO  THEORY  OF  STRUCTURES. 

Corollary.  —  From  the  last  equation,  —^  =    —  ,  and  -y- 

is  consequently  a  measure  of  the  longitudinal  stiffness  of  a  bar, 
so  that  for  the  same  material,  the  stiffness  varies  directly  as  the 
sectional  area  and  inversely  as  the  length,  while  for  different 
materials  it  also  varies  directly  as  the  coefficient  of  elasticity. 
Work  of  Extension.  —  The  force  producing  the  increment  dx 

has    for    its    least   value   P(=£A^j,  for  its  greatest  value 
P-\-dP,  and  for  its  mean  value  P-\  --  ,  so  that  the  work  done 

is  fP-\  --  \dx  =  Pdx,  approximately. 

Hence  the  work  done  in  stretching  the  bar  until  its  length 
is  L  +  /is  equal  to 


24.  On  the  Oscillatory  Motion  of  a  Weight  at  the  End 
o         of  a  Vertical  Elastic  Rod.  —  An  elastic  rod  of  natu- 
*    ral  length  L(OA)  and  sectional  area  A  is  suspended 
from  O,  and  carries  a  weight  P  at  its  lower  end,  which 
elongates  the  rod  until  its  length  is  OB  =  L  -f-  /. 

Assume  that    the  mass  of   the  rod  as   compared 
with  P  is  sufficiently  small  to  be  disregarded,  then 

P=EAl-L. 

A 

If  the  weight  is  made  to  descend  to  a  point  C,  and 
|  is  then  left  free  to  return  to  its  state  of  equilibrium,  it 
!Q  must  necessarily  describe  a  series  of  vertical  oscilla- 

tions about  B  as  centre. 
FIG.  196.          Take  g  as  the  orjgin)  and  at  any  time  t  iet  t|ie 

weight  be  at  M  distant  x  from  B  ;  also  let  BC  —  c. 

Two  cases  may  be  considered. 

First,  suppose  the  end  of   the  rod  to  be  gradiially  forced 
down  to  C  and  then  suddenly  released. 


OSCILLA  TOR  Y  MO  TION.  i 

According  to  the  principle  of  the  conservation  of  energy, 

( -7- )    =  the  work  done  between  C  and  M 

2  \dtl 


g 

_EA 

or 

Pi  (dxV      Pi 


and  hence 

z>,  the  velocity  of  the  weight  at  M,  =  \     —(?  —  x*\\ 

V  / 

Now  v  is  zero  when  x  =  ±  c,  so  that  the  weight  will  rise 
above  B  to  a  point  C,  where  BCl  ,=•=.;  =  .#£7. 
Again,  from  the  last  equation, 


.  /€      __^£_ 

V7  ~(V-^)* 


and  integrating  between  the  limits  o  and  x, 


and  the  oscillations  are  therefore  isochronous. 
When  .*  =  c, 

•x 


and  the  time  of  a  complete  oscillation  is 


g 


IQ2  THEORY  OF  STRUCTURES. 

Next,  suppose  the  oscillatory  motion  to  be  caused  by  a 
weight  P  falling  without  friction  from  a  point  D,  and  being 
suddenly  checked  and  held  by  a  catch  at  the  lower  end  of  the 
rod. 

Take  the  same  origin  and  data  as  before,  and  let  AD  =  h. 

The  elastic  resistance  of  the  rod  at  the  time  t  is 


L 
and  the  equation  of  motion  of  the  weight  is 

Pcfx 


~  g 

or      =  - 


Integrating, 

!dx\?  g 

!  —  j   =  —  jx*  -f-  cl  ,  cl  being  a  constant  of  integration. 


dx  P~ 

But  -7-  is  zero  when  x  =  c,  and  c.  =^c*. 
at  I 


Hence 


This  is  precisely  the  same  equation  as  was  obtained  in  the 
first  case,  and  between  the  limits  o  and  x 


rg    .  ,* 
yf  :  sm  7- 


OSCILLATORY  MOTION.  1 93 

so  that  the  motion  is  isochronous,  and  the  time  of  a  complete 
oscillation  is 


•x4 


Cor.  i.  When  x  =  —  I, 


and  hence 

f  (<>-/<)  = 
or 


Cor.  2.  If  h  =  o,  i.e.,  if  the  weight  is  merely  placed  upon 
the  rod  at  the  end  A,  c  =  ±  /,  and  the  amplitude  of  the 
oscillation  is  twice  the  statical  elongation  due  to  P. 

Cor.  3.  The  rod  may  be  safely  stretched  until  its  length  is 
L  +  /,  while  a  further  elongation  c  might  prove  most  injurious 
to  its  elasticity,  which  shows  the  detrimental  effect  of  vibratory 
motion.  If  a  small  downward  force  Q  is  applied  to  Pwhen  it 
has  reached  the  end  of  its  vibration,  it  will  produce  a  corre- 
sponding descent,  and  the  weight  P  will  then  ascend  an  equal 
distance  above  its  neutral  position.  At  the  end  of  the  interval 
corresponding  to  P's  natural  period  of  vibration,  apply  the 
force  again,  and  P  will  descend  still  further.  This  process 
may  be  continued  indefinitely,  until  at  last  rupture  takes  place, 
however  small  Pand  Q  may  be.  If  Q  is  applied  at  irregular 
intervals,  the  amplitude  of  the  oscillations  will  still  be  increased, 
but  the  increase  will  be  followed  by  a  decrease,  and  so  on  con- 
tinually. In  practice  the  problem  becomes  much  more  com- 
plex on  account  of  local  conditions,  but  experience  shows  that 
a  fluctuation  of  stress  is  always  more  injurious  to  a  structure 
than  the  stress  due  to  the  maximum  load,  and  that  the  injury 


194  THEORY  OF  STRUCTURES. 

is  aggravated  as  the  periods  of  fluctuation  and  of  vibration  of 
the  structure  become  more  nearly  synchronous. 

An  example  of  a  fluctuating  load  is  a  procession  marching 
in  time  across  a  suspension-bridge,  which  may  strain  it  far 
more  severely  than  a  much  greater  dead  load,  and  may  set  up 
a  synchronous  vibration  which  may  prove  absolutely  dangerous. 
In  fact,  a  bridge  has  been  known  to  fail  from  this  cause. 

Cor.  4.  The  coefficient  of  elasticity  of  the  rod  may  be  ap- 
proximately found  by  means  of  the  formula 


/ 

—  y 
g 


T  being  the  time  of  a  complete  oscillation.     For  suppose  that 
the  rod  emits  a  musical  note  of  n  vibrations  per  second,  then 


g 

is  the  time  of  travel  from  C  to  Cl ; 

n- 

and  hence 


Cor.  5.  Suppose  that  the  weight  is  perfectly  free  to  slide 
along  the  rod.  When  it  returns  to  A,  it  will  leave  the  end  of 
the  rod  and  rise  with  a  certain  initial  velocity.  This  velocity 
is  evidently  V2gKt  and  the  weight  accordingly  ascends  to  D, 
then  falls  again,  repeats  the  former  operation,  and  so  on.  The 
equations  of  motion  are  in  this  case  only  true  for  values  of  x 
between  x  —  +  c  and  x  =  —  /. 

25.  On  the  Oscillatory  Motion  of  a  Weight  at  the  End 
of  a  Vertical  Elastic  Rod  of  Appreciable  Mass. — Suppose 
the  mass  of  the  rod  to  be  taken  into  account,  and  assume : 

(a)  That  all  the  particles  of  the  rod  move  in  directions  par- 
allel to  the  axis  of  the  rod, 


OSCILLATORY  MOTION.  1 95 

(b)  That  all  the  particles,  which  at  any  instant  are  in  a  plane 
perpendicular  to  the  axis,  remain  in  that  plane  at  all  times. 

As  before,  the  rod  OA  of  natural  length  L  and  sectional 
area  A  is  fixed  at  O  and  carries  a  weight  Pl  at  A. 

Take  O  as  the  origin,  and  let  OX  be  the  axis  of  the  rod. 

Let  £,£-)-  d£,  and  x,  x  -j-  dx,  be  respectively  the  actual 
and  natural  distances  from  O  of  the  two  consecutive 
sections  MM,  M'M'.  o 

Let  p0  be  the  natural  density  of  the  rod,  and  p 
the  density  of  the  section  MM,  distant  £  from  O. 

The  forces  which  act  upon  the  rod  are : 

(a)  The  upward  and  constant  force  P0  at  0.  M> 

(b)  The  weight  Pl  at  A. 

(c)  The  weight  of  the  rod. 

(d)  A  force  X  per  unit  of  mass  through  the  slice 
bounded  by  the  planes  MM,  M' M',  distant  %  and 
B>  -\-  d£>,  respectively,  from  O. 

Suppose  the  rod,  after  equilibrium  has  been  es- 
tablished, to  be  cut  at  the  plane  M'M'.     In  order  to 
maintain  the  equilibrium  of  the  portion  OM ' M'  it 
will  be  necessary  to  apply  to  the  surface  of  this  plane  a  certain 
force  P,  and  the  equation  of  equilibrium  becomes 


=  o. 


But  if  the  thickness  d$  of  the  slice  MM'  is  indefinitely 
diminished,  P  is  evidently  the  elastic  reaction,  and  its  value  is 


Hence 


=  o. 


IQ  THEORY  OF  STRUCTURES. 

Differentiating  with  respect  to  x, 


~Kvt  pdS  —  ft^dx, 


~  +  p,gA  =  o, 


or  - 


Also,  pnAXdx  is  the  resistance  to  acceleration  arising  from 
the  inertia  of  the  slice,  and  is  therefore  equal  to 


so  that 


•  df 

Hence 


To  solve  this  equation. — In  the  state  of  equilibrium, 

'*•£-') 

is   the   tension    in    the    section    of   which   the   distance   from 


OSCILLATORY  MOTION.  197 

0  is  x,  and  counterbalances  the  weight  Pl  and   the  weight 
pQA(l  —  x)g  of  the  portion  AMN  of  the  rod. 


.\EA         ~  »   - 


or 


Integrating, 

+    *-   ----  c> 


There  is  no  constant  of  integration,  as  x  and  %  vanish 
together. 

This  value  of  ^  is  a  particular  solution  of  (i),  and  is  inde- 
pendent of  /. 


Put        5  =  * 

z  being  a  new  function  of  x  and  t.     Then 


_     _       p.  z 

*''     ~g+*'     and 


Hence,  from  eq.  (i), 


d*z       Ed*z  d*z  E 

—75  =  ~  T~^  =  v\  3~T  >    where     v?  =  —  , 
d?       p.dx*         l  dx*  p0 


The  integral  of  this  equation  is  of  the  form 


IQ8  THEORY  OF  STRUCTURES. 


=  A  /  —  )  being  the  velocity  of  propagation  of  the  vibrations. 
The  full  solution  of  (i)  is  therefore  of  the  form 

p, 


26.  Inertia  —  Balancing.—  Newton's  First  Law  of  Motion, 
called  also  the  Law  of  Inertia,  states  that"  a  body  will  continue 
in  a  state  of  rest  or  of  uniform  motion  in  a  straight  line  unless 
it  is  made  to  change  that  state  by  external  forces." 

This  property  of  resisting  a  change  of  state  is  termed 
inertia,  and  in  dynamics  is  always  employed  to  measure  the 
quantity  of  matter  contained  in  a  body,  i.e.,  its  mass,  to 
which  the  inertia  must  be  necessarily  proportional.  Thus,  to 
induce  motion  in  a  body,  energy  must  be  expended,  and  must 
again  be  absorbed  before  it  can  be  brought  to  rest.  The  inertia 
of  the  reciprocating  parts  of  a  machine  may  therefore  heavily 
strain  the  framework,  which  should  be  bolted  to  a  firm  foun- 
dation, or  must  be  sufficiently  massive  to  counteract  by  its 
weight  the  otherwise  unbalanced  forces. 

EXAMPLE  I.  Consider  the  case  of  a  direct-acting  horizontal 

steam-engine,  Fig.  198.  At  any 
given  instant  let  the  crank  OP 
and  the  connecting-rod  CP  make 
angles  0  and  0,  respectively,  with 
the  line  of  stroke  AB. 

Let  v  be   the  velocity  of   the 
FIG.  198.  crank-pin  centre  P,  and  let  u  be 

the  corresponding  piston  velocity,  which  must  evidently  be  the 
same  as  that  of  the  end  C  of  the  connecting-rod. 

Let  OP  produced  meet  the  vertical  through  C  in  /. 
At  the  moment  under  consideration,  the  points  C  and  Pare 
turning  about  /  as  an  instantaneous  centre. 


1C      sin 

IP  ~  COS  0 


INpR  TIA  —BA  LA  NCING. 

Let  W  be  the  weight  of  the  reciprocating  parts,  i.e.,  the 
piston-head,  piston-rod,  cross-head  (or  motion-block),  and  a  por- 
tion of  the  connecting-rod. 

Assume  (i)  that  the  motion  of  the  crank-pin  centre  is  uni- 
form ; 

(2)  that  the  obliquity  of  the  connecting-rod  may 
be  disregarded  without  sensible  error,  and 
/.  0  —  o. 

Draw  PN  perpendicular  to  AB,  and  let  ON=.x\  ON  is 
equal  to  the  distance   of  the  piston  from   the  centre  of  the 
stroke,  corresponding  to  the  position  OP  of  the  crank. 
The  kinetic  energy  of  the  reciprocating  parts 

W  21*       W  v*  sin8  0       W  V*  i        x^ 
g    2  ""  g        2          "  g    2  \J    "  r* 

r  being  the  radius  OP. 

.'.  the  change  of  kinetic  energy,  or  work  done,  corresponding 
to  the  values  xl ,  x^  of  x, 


W^<x;-x?\ 

~     r    2  r2        )' 


g    2 

Let  R  be  the  mean  pressure  which,  acting  during  the  same 
interval,  would  dp  the  same  work.     Then 

W  v*  x?  -  x? 

7  T~^~    =*(*'-*•>' 

and 


Hence,  in    the  limit,  when  the  interval   is  indefinitely  small, 
#,  =  x^  =  x,  and  the  pressure  corresponding  to  x  becomes 

wv 

R  = -r-ar. 

' 


200  THEORY   OF  STRUCTURES. 

This  is  the  pressure  due  to  inertia,  and  may  be  written  in  the 
form 


R-  C 
-67> 

/       W  v*\ 

C  [•= 1  being  the  centrifugal  force  of  W  assumed  con- 
centrated at  the  crank-pin  centre.  R  is  a  maximum  and  equal 
to  C  when  x  =  r,  i.e.,  at  the  points  A,  B,  and  its  value  at 
intermediate  points  may  be  represented  by  the  vertical  ordi- 
nates  to  AB  from  the  straight  line  EOF  drawn  so  that 
AE  =  BF  =  C.  In  low-speed  engines,  C  may  be  so  small  that 
the  effect  of  inertia  may  be  disregarded,  but  in  quick-running 
engines,  C  may  become  very  large  and  the  inertia  of  the  recip- 
rocating parts  may  give  rise  to  excessive  strains. 

Another  force  acting  upon  the  crank-shaft  is  the  centrifu- 
gal force  of  the  crank,  crank-pin,  and  of  that  portion  of  the  con- 
necting-rod which  may  be  supposed  to  rotate  with  the  crank- 
pin. 

Let  w  be  the  weight  of  the  mass  concentrated  at  the  crank- 
pin  centre  which  will  produce  the  same  centrifugal  force  as 
these  rotating  pieces  (i.e.,  wr  —  sum  of  products  of  the  weights 
of  the  several  pieces  into  the  distances  of  their  centres  of  gravity 
from  O). 

TU  -r  i    r  e  W   V* 

The  centrifugal  force  of  w  = . 

g    * 

Thus  the  total  maximum  pressure  on  the  crank-shaft 


A  being  the  uniform  angular  velocity  of  the  crank-pin. 

This  pressure  may  be  counteracted  by  placing  a  suitable 
balance-weight  (or  weights)  in  such  a  position  as  to  develop  in 
the  opposite  direction  a  centrifugal  force  of  equal  magnitude. 


INER  TIA  —BA  LA  NCING.  20  1 

Let  Wl  be  such  a  weight  and  R  its  distance  from  O.    Then 


or 


from  which,  if  R  is  given,  Wl  may  be  obtained. 

During  the  first  half  of  the  stroke  an  amount  of  energy 
represented  by  the  triangle  AEO  is  absorbed  in  accelerating  the 
reciprocating  parts,  and  the  same  amount,  represented  by  the 
triangle  EOF,  is  given  out  during  the  second  half  of  the  stroke 
when  the  reciprocating  parts  are  being  retarded. 

During  the  up-stroke  of  a  vertical  engine  the  weights  of  the 
reciprocating  parts  act  in  a  direction  opposite  to  the  motion  of 
the  piston,  while  during  the  down-stroke  they  act  in  the  same 
direction. 

lr\AE  produced  (Fig.  199)  take  EE'  to  represent  the  weight 
of  the  reciprocating  parts  on  the  same  scale 
as  AE  represents  the  pressure  due  to  inertia.  E 
Draw  E'O'F'  parallel  to  EOF.  A  I 

During   the    up-stroke  the  ordinates   of 
E'O'  represent  the  pressures  required  to  ac- 
celerate the  reciprocating  parts,  the  pressures  while  they  are 
retarded  being  represented  by  the  ordinates  of  O'  F'  . 

The  case  is  exactly  reversed  in  the  down-stroke. 

N.B.  —  The  formula  R  =  C—  may  be  easily  deduced  as 
follows  : 

du  dd      v* 

&  =  ^  sin  6;  the  acceleration  =  -77  =  v  cos  ft  -y-  —  -;#  ; 

at  at      r 

W  du 

.*.  --  -j-  =  accelerating  force  =  force  due  to  inertia 
g   at 

W  v*  * 

=  --  TX  =  C~. 

g  r*  r 


202  THEORY  OF  STRUCTURES. 

Ex.  2.  Consider  a  double-cylinder  engine  with  two  cranks 
at  right  angles,  and  let  d  be  the  distance  between  the  centre 
lines  of  the  cylinders  (Fig.  2OO). 


of  Cylr. 


Centre  line 


±C.sin  d 
FIG.  200. 

The  pressures  due  to  inertia  transmitted  to  the  crank-pins 
when  one  of  the  cranks  makes  an  angle  6  with  the  line  of 
stroke  are 

P,=  C  cos  0     and     P,  =  C  sin  V. 
These  are  equivalent  to  a  single  alternating  force 
P  =  C(cos  6  ±  sin  0) 

acting  half-way  between  the  lines  of  stroke,  together  with  a 
couple  of  moment 

M  =  P-  =  C-(cos  0  ±  sin  0). 

2         •    2^ 

The  force  and  couple  are  twice  reversed  in  each  revolution, 
and  their  maximum  values  are 

Cd 

P,naX.  =  CV2     and      Mmax,  =  —  Y2. 

In  order  to  avoid  the  evils  that  might  result  from  the  action 
of  the  force  and  couple  at  high  speeds,  suitable  weights  are 
introduced  in  such  positions  that  the  centrifugal  forces  due  to 


INER  TIA  —BA  LA  NCING.  203 

their  rotation  tend  to  balance  both  the  force  and  the  couple. 
For  example,  the   weights  may  be  placed 
upon  the  fly-wheels,  or  again,  upon  the  driv- 
ing-wheels of  a  locomotive. 

Let  a  balance-weight  Q  be  placed  nearly 
diametrically  opposite  to  the  centre  of  each 
crank-pin  (Fig.  201),  and  let  R  be  the  distance 
from  the  axis  to  the  centre  of  gravity  of  Q. 

Let  e  be  the  horizontal  distance  between  FlG>  20I> 

the  balance-weights. 

The  centrifugal  force  /^due  to  the  rotation  of  Q 

g (velocity  of  Q}"      Q  I  R*_  ,  _  QR   a 
R  ~ g    ~Rr*V    ~  gr*V> 

and  this  force  F  is  equivalent  to  a  single  force  F acting  half-way 
between  the  weights  and  to  a  couple  of  moment 

^ 
F  — .     Let  0  be  the  angle  between  the  radius 

to  a  balance-weight,  and  the  common  bisector 
of  the  angle  between  the  two  cranks  (Fig.  202). 
Since  there  are  two  weights  Q,  there  will 

€ 

FIG.  202.  be  two  couples  each  of  moment  F  —  ,  and  two 

forces  each  equal  to  F  acting  half-way  between  the  weights, 
the  angle  between  the  axes  of  the  couples  being  180°—  20,  and 
that  between  the  forces  being  20.  The  moment  of  the  result- 
ant couple  is  Fe  sin  0,  and  its  axis  bisects  the  angle  between 
the  axes  of  the  separate  couples;  the  resultant  force  parallel 
to  the  line  of  stroke  —  2.F  cos  0. 

Q  and  0  may  now  be  chosen  so  that 

2/rcos  0  =  maximum   alternating  force  =  C  ^2, 
and 

Fe  sin  0  =  maximum  alternating  couple  =  —  1/2. 

d 

.'.  tan  0  —  -  , 


204  THEORY  OF  STRUCTURES. 

and 


or 


_A  /?a  +  *ra 
g~?"    ~  g  r 

and 

W 


Ex.  3.  Again,  the   pressure    C  at   a   dead   point  may  be 
balanced  by  a  weight  Q  diametrically  opposite. 
If  R  is  the  radius  of  the  weight-circle,  then 


Wv*  QR  a 

and 

R 

e  -\-  d 
The  weight  Q  may  be  replaced  by  a  weight  Q on  the 

near  and   a  weight  Q on  the  far  wheel.     Thus,  since  the 

cranks  are  at  right  angles,  there  will  be  two  weights  90°  apart 
on  each  wheel,  viz.,  Q—  -  in  line  with  the  crank  and  Q . 

These  two  weights,  again,  may  be  replaced  by  a  single  weight 
B  whose  centrifugal  force  is  the  resultant  of  the  centrifugal 
forces  of  the  two  weights.  Thus 

er-c*  -c/^ 
^ 


B  </_Y_  (Q  e  +  d  z/V       (Q  e-d  z/V 
R)"\fr     2e      RJ     "  V      2e      Rr 


vf  being  the  linear  velocity  at  the  circumference  of  the  weight- 
circle. 


2e 


CURVES  OF  PISTON    VELOCITY. 


205 


or 


B=V 


If  a  is  the  angle  between  the  radius  to  the  greater  weight 

us, 
Qe-dv' 


e-\-d 
Q and  the  crank  radius, 


tan  a  - 

' 


'* 
R       e~ 


Note.  —  In  outside-cylinder  engines  e  —  d  is  approximately 
nil,  and  B  =  Q  =  W~. 

/V 

27.  Curves  of  Piston  Velocity.  —  Consider  the  engine  in 
Ex.  i. 

a 


FIG.  203. 

Let  CP  produced  intersect  the  vertical  through  0  in  T,  and 
in  OPtakeOT'  =  OT. 

The  piston  velocity  u  and  the  velocity  V  of  the  crank-pin 
centre  are  connected  by  the  relation 


sin(fl4-0)__<9:r 

cos  0        ~~OP~~~OP' 


.     .     .     (I) 


If  the  velocity  v  is  assumed  constant,  and  if  it  is  represented 
by  OP,  then  on  the  same  scale  OT'  will  represent  the  piston 
velocity  u.  Drawing  similar  lines  to  represent  the  value  of  u 


2O6 


THEORY  OF  STRUCTURES. 


for  every  position  of  the  crank,  the  locus  of  T'  will  be  found  to 
consist  of  two  closed  curves  OGS,  OUT,  called  the  polar  curves 
of  piston  velocity.  They  pass  through  the  point  O  and  through 
the  ends  5  and  T  of  the  vertical  diameter.  On  the  side  towards 
the  cylinder  they  lie  outside  the  circles  having  OS  and  OT  as 
diameters,  while  on  the  side  away  from  the  cylinder  they  lie 
inside  the  circles.  If  the  connecting-rod  is  so  long  that  its 
obliquity  may  be  disregarded, 


=  o     and 


u  = 


0, 


and  the  curves  coincide  with  the  circles. 

A  rectangular  diagram  of  velocity  may  be  drawn  as  follows  : 


C    N 

FIG.  204. 


C      M 


Upon  the  vertical  through  C,  Fig.  204,  take  CL  =  OT\  the 
locus  of  L  is  the  curve  required  for  one  stroke.  A  similar 
curve  may  be  drawn  for  the  return  stroke  either  below  MN  or 
upon  the  prolongation  NR  (=  MN)  of  MN. 

If  the  obliquity  of  the  connecting-rod  is  neglected,  the 
curves  evidently  coincide  with  the  semicircles  upon  MN  and 
NR,  MN  (=  NR)  defining  the  extreme  positions  of  C.  The 
obliquity,  however,  causes  the  actual  curve  to  fall  above  the 
semicircle  during  the  first  half  of  the  stroke,  and  below  during 
the  second  half. 

Again,  let  the  connecting-rod  (/)  =  n  cranks  (r).    Then 

sin  B       I 

— —  —  —  z^i  n, 

sin  0       r 

and  by  eq.  I, 

u  =  v  (sin  0  -\-  cos  8  tan  0)  =  v  (  sin  8  -| -.  ).     (2) 

\  #V  —  sin2  0/ 


CURVE   OF  CRANK-EFFORT—CURVES  OF  ENERGY.       2O? 

If  the  obliquity  is  very  small, 


and 


sin  0 

tan  0  =  sm  0  = ,  approximately, 


sin  #cos  0 


sin20\ 

—  ). 


28.  Curve  of  Crank-effort.  —  The  crank-effort  F  for  any 
position  OP  of  the  crank  is  the  component  along  the  tangent 
at  P  of  the  thrust  along  the  connecting-rod. 


This  thrust  = 


cos  <p' 


COS  0 


If  the  pressure  P  upon  the  piston  is  constant,  and  if  it  is  rep- 
resented by  OP,  then,  on  the  same  scale,  OT' ,  Fig.  203,  will  rep- 
resent the  crank-effort.  Thus,  the  curves  of  piston  velocity 
already  drawn  may  also  be  taken 
to  represent  curves  of  crank-effort. 
If  the  pressure  P  is  variable,  as  is 
usually  the  case,  let  OP,  the  crank 
radius,  represent  the  initial  value 
of  P.  After  expansion  has  begun, 
take  OP'  in  OP,  for  any  position  OP 
of  the  crank,  to  represent  the  cor- 
responding pressure  which  may  be 
directly  obtained  from  the  indicator-diagram.  Draw  P'T' 
parallel  to  PT,  and  take  OT"  =  OT .  Then  OT"  will  repre- 
sent the  required  crank-effort,  and  the  linear  and  polar  diagrams 
may  be  drawn  as  already  described. 

29.  Curves  of  Energy— Fluctuation  of  Energy. — In  the 
curve  of  crank-effort  as  usually  drawn,  the  crank-effort  for  any 
position  OP  oi  the  crank  is  the  ordinate  S'H,  the  abscissa  DH 
being  equal  to  the  arc  AP,  i.e.,  to  the  distance  traversed  by  the 


FIG.  205. 


208 


THEORY  OF  STRUCTURES. 


point  of  application  of  the  crank-effort.     Thus,  DSE  andEVG 
being  the  curves, 

DE  =  EG  =  semi-circumference  of  crank-circle  —  ?rr. 
If  the  obliquity  is  neglected,  the  curves  of  crank-effort  are 
the  two  curves  of  sines  shown  by  the  dotted  lines. 


H 


The  area  DS'H  also  evidently  represents  the  work  done  as 
the  crank  moves  from  OA  to  OP,  and  the  total  work  done  is 
represented  by  the  area  DSE  in  the  forward  and  by  EVG  in 
the  return  stroke. 

Let  F9  be  the  mean  crank-effort.     Then 


F0  X  2nr  —  2PX  2r, 
assuming  Pto  be  constant. 


2p 
Draw  the  horizontal  line  1234567  at  the  distance  —  from 

7t 

DEG,  and  intersecting  the  verticals  through  D,  E,  and  G  in  i, 
4,  and  7,  and  the  curves  in  2,  3,  5,  and  6.  The  engine  may  be 
supposed  to  work  against  a  constant  resistance  R  equal  and 
opposite  to  the  mean  crank-effort  F0 . 

From  D  to  2,  R  >  crank-effort,  and  the  speed  must  there- 
fore continually  diminish. 

From  2  to  3,  R  <  crank-effort,  and  the  speed  must  contin- 
ually increase. 

Thus  2  is  a  point  of  min.  velocity,  and  therefore  also  of 
min.  kinetic  energy. 

From  3  to  E,  R  >  crank-effort,  and  the  speed  must  contin- 
ually diminish. 


CURVES   OF  ENERGY—  FLUCTUATION   OF  ENERGY.     2OQ 

Thus  3  is  a  point  of  max.  velocity,  and  therefore  also  of 
max.  kinetic  energy. 

Similarly,  in  the  return  stroke,  5  and  6  are  points  of  min. 
and  max.  velocity,  respectively. 

The  change  or  fluctuation  of  kinetic  energy  from  2  to  3  — 
area  283,  bounded  by  the  curve  and  by  23. 

The  fluctuation  from  3  to  5  =  area  3^5,  bounded  by  35  and 
by  the  curve. 

F      Fr 
Again,  since  —  —  —  -,  the  ordinates  of  the  curves  may  be 

taken  to  represent  the  moments  of  crank-effort,  and  the  abscissae 
are  then  the  corresponding  values  of  0. 

The  work  done  between  A  and  any  other  position  P  of 
the  crank-pin 


=  Pr(i  —  cos  0  +  n  —  \'n*  —  si 


If  there  are  two  or  more  cranks,  the  ordinates  of  the  crank- 
effort  curve  will  be  equal  to  the  algebraic  sums  of  the  several 
crank-efforts.  For  example,  if  the  two  cranks  are  at  right 
angles,  and  if  Flt  F2  are  the  crank-efforts  when  one  of  the 
cranks  (F^  makes  an  angle  6  with  the  line  of  stroke, 


sn 


and 


.-.  Ft-{-F9  =  P(sin  0  +  cos  0)  =  combined  crank-effort, 

P  being  supposed  constant. 

Note.  —  In  the  case  of  the  polar  curves  of  crank-effort,  if  a 
circle  is  described  with  O  as  centre  and  a  radius  =  mean  crank- 

2p 
effort  =  -—  ,  it  will  intersect  the  curves  in  four  points,  which 

are  necessarily  points  of  max.  and  min.  velocity. 


210 


THEORY  OF  STRUCTURES. 


U3 
ffl 
IN 


§.§,!§ 

0    0    o'  « 


c  a 

3-0 

o  o 


o  o  -<t- 
riTrM 


OO    ro  (N   O>  O 


»-<«  a 


000300000300   ro-i-b 
C         CCCCCCCCC 


§888 
o_  q  q_  q_ 


Com- 
pressio 


8  8 

0  VO 

is 


vo~      vo"      (xo   rn  cT  -4-  10  to  <>  -r"-»-<? 

*O       vovovOioincKoooooo  vo        o 


6  : 

rt 


~    «     ::::::::  .-  : 


si  °    ;.   :  :'CE-S^« 

ti  1     sl    =   «lin 

:-i  liliiliillil 

s 


^ 


2  «  S-a  J>.>  ^ 

lig  4 


li  I 

CQ         cq 


1 


TABLES. 


211 


THE   STRENGTHS,    ELASTICITIES,    AND   WEIGHTS   OF 
VARIOUS   ALLOYS,    ETC. 


Material. 

Max.  Load  on  Original  Area 
in  Ibs.  per  sq.  in. 

Young's 
Modulus, 

(in  Ibs.). 

Coef- 
ficient of 
Rigidity, 
G 

(in  Ibs.). 

Weight 
in  Ibs. 
per  cu.  ft. 

Tension. 

Com- 
pression. 

Shear- 
ing. 

A  iuminum                  

• 
28,800 
17,600 

52,000 

30,000 
60,000 
36,000 
1,850 
3,100 
57,000 
4,980 
7,5oo 
4,000 

10,380 

58,000 
7,100 

9,600.000 
9,100,000 
15,000,000 
14,000,000 
15,000.000 
15,000,000 
17,000,000 
9,900,000 
710,000 
996,000 
14,000,000 
5,690,000 
13,500,000 
25,000 

3,600,000 
3.400,000 
5,600,000 

5,700,000 
5,700,000 

5,250,000 
2,140,000 

160  to  166 
487  to  524.4 

$ 

555 
529 
712 

456  to  468 
424  to  449 

1  '      (hammered)  

Brass  wire 

Copper-plate,  hammered  — 
annealed  

Gun-metal  

Lead                                      .     . 

Phosphor-bronze        

Tin 

Zinc  
Leather 

THE  STRENGTHS,  ELASTICITIES,  AND  WEIGHTS  OF  TIMBERS. 

This  table  contains  the  results  of  the  most  recent  and 
most  reliable  experiments,  but,  generally  speaking,  only  small 
specimens  of  the  material  have  been  tested.  It  is  found  that 
the  strength,  elasticity,  and  weight  of  a  timber  are  affected  by 
the  soil,  age,  seasoning,  per  cent  of  moisture,  position  in  the 
log,  etc.,  and  hence  it  is  not  surprising  that  specimens  even 
when  cut  out  of  the  same  log  show  results  which  often  differ 
very  widely  from  the  mean.  Additional  experiments  on  large 
timbers  are  needed,  and  in  each  case  should  be  accompanied 
by  a  complete  history  of  the  specimen  from  the  time  of  felling. 


Description  of 
Timber. 

Tensile 
Strength 
in  tons 
per  sq.  in. 

Com- 
pressive 
Strength 
in  tons 
per  sq.  in. 
along 
Fibres. 

Shearing 
Strength 
in  tons 
per  sq.  in. 
along 
Fibres. 

Young's 
Modulus, 
E 
(in  tons). 

Coef- 
ficient 
of 
Rigid- 

*% 

Coefficient 
of  Bending 
Strength 
in  tons 
per  sq.  in. 

Weight 
in  Ibs. 
per 
cu.  ft. 

Acacia 

Alder  

4.5  to  6.3 

3-  J 

Apple 

88 

Ash,  Canadian.. 
Ash,  English  .  .. 
Beech 

2-45 
5.35107.58 
4  9  to  9  8 

2-5 

3467 

.2  tO   .312 

620 
723 
607 

47 
43  to  53 

Birch 

6  60 

Box  " 

4.6 

807 

Blue  gum... 

2.  7 

3.078 

509 

c  86 

Cedar  
Chestnut  
Ebony  
Elm,  Canadian.. 

2.23104.9 
4-3 

2.56 

8.48 
2.9 

.308 

217 
509 

IIOO 

35  to  47 
35  to  41 

47 

212 


THEORY   OF  STRUCTURES. 


THE    STRENGTHS,    ELASTICITIES,    AND    WEIGHTS    OF   TIMBERS 

(Continued.} 


Description  of 
Timber. 

Tensile 
Strength 
in  tons 
per  sq.  in. 

Com- 
pressive 
Strength 
in  tons 
per  sq.  in. 
along 
Fibres. 

Shearing 
Strength 
in  tons 
per  sq.  in. 
along 
Fibres. 

Young's 
Mod-lus, 
E 

(in  tons). 

Coef- 
ficient 
of 
Rigid- 

Coefficient 
of  Bending 
Strength 
in  tons 
per  sq.  in. 

Weight 
in  Ibs. 
per 
cu.  ft. 

Elm,  Eng  

5-89 

4.6 

* 

34  to  3; 

Green  heart  ... 

2.7  to  4.1 

4.4610  6.5 

759 

58  to  72 

Hawthorn  

4.68 

Hazel   

8.48 

Hornbeam   

9.1 

2.6 

47* 

Iron  bark    

7.12 

4-54 

8.15 

Ironwood  

4-31 

5-21 

Jarrah  

3-2 

4-J3 

Lancewood  

3.6  to  6.7 

42  to  63 

Larch  

3.92  to  4.55 

1.42  to  2.45 

32  to  38 

Lignum  vitae.  .  .. 

5-26 

4.46 

446 

7.18 

41  to  83 

Locust        

4-5to6.7 

i-33 

•535 

Mahogany,  Span.. 

i.  7  to  7.  3 

3-3 

3-3 

560  to  1339 

53 

''          Hond 

1.3  to  3.6 

712  to  879 

35 

Maple  

4.7  to  7  .  7 

2  23 

N  J 

49 

Mora  

4-i 

4-4 

830 

II 

57  to  68 

*0ak.  Am  

577 

N- 

2    712 

Oak,  Am   red.  .. 

4-46 

1.89  to  2.6 

.324  to  .446 

.* 

61 

"    white 

8.8 

2.84 

•335  to  .431 

tt 

4  •  55 

61 

"     Eng  

5-4 

4-4 

664 

* 

49  to  58 

Pine,  Dantzic.  .. 

3-5 

1025 

3 

36 

"     Memel  

4-5 

ij 

34 

"     pitch  

4.6 

3-5 

| 

41  to  58 

*  "     red  

670 

<*_ 

2-93 

34 

"     red  

1.7  to  6.67 

2.  4  to  3 

962 

5, 

34 

*  "     yellow  

779 

7* 

3-255 

41     yellow.  .  .. 

2.2  tO  6.87 

2.4  to  3.6 

.227 

900 

4-Si 

32 

*  "     white  

484 

2.146 

"     white  .  . 

1.3  to  5.1 

2.24 

.  119  to  .  164 

3-03 

3° 

Plane 

5  •  4 

604 

40 

Poplar  

2-94 

i  8 

34° 

23  to  26 

*  Spvuce 

594 

2.18 

2.99  to  5.97 

438  to  737 

1.63  to  2.86 

n 

.  113  to  .  167 

700 

29  to  32 

Sycamore  
Teak      .. 

5-8 
4.7  to  6.7 

3-16 
5-35 

464 

IOOO 

36  to  43 
41  to  52 

Walnut  

3-5 

2-7 

38  to  57 

Willow  

4.6  to  6.25 

629 

24  to  35 

*The  results  for  these  timbers  are  deduced  from  experiments  carried  out  by  Bauschinger 
Lanza,  and  others,  on  comparatively  large  specimens. 

THE    BREAKING    WEIGHTS    AND    COEFFICIENTS    OF    BENDING 
STRENGTH     IN    TOMS    (OF    2240    LBS.)    OF    VARIOUS     RECT- 
ANGULAR   BEAMS,    THE    WEIGHTS    HAVING 
BEEN    UNIFORMLY    DISTRIBUTED. 


Material. 

Clear 
Span 
between 
Supports 
in  inches. 

Breadth 
in 
inches. 

Depth 
in 
inches. 

Mean 
Breaking 
Weight  of 
each  Joist 
or  Beam. 

Coef- 
ficient of 
Bending 
Strength. 

g 

5.66 

.48 

142 

3* 

II 

7.89. 

.08 

"          «*             '*           2  beams        

126 

14 

15 

60.97 

.83 

126 

46  6 

.6 

142 

3i 

II 

8.29 

.08 

g 

5.7 

•49 

Pine  (Baltic),  2  beams  
Baltic  redwood  deal  (Wyberg),  2  joists  ... 
Spruce  deals  (St.  John),  3  pairs  with  bridg- 
ing oieces.  .  .            

126 
142 

142 

'Si 
3 

3 

*3* 
9 

9 

58-43 
5-75 

6.81 

.24 
•52 

2.98 

TABLES. 


213 


THE    BREAKING    WEIGHTS    AND    COEFFICIENTS    OF    BENDING 
STRENGTH    IN    TONS    (OF    2240    LBS.)    OF     VARIOUS    RECT- 
ANGULAR BEAMS  LOADED  AT  THE  CENTRE. 


Material. 

Clear 
Span 
between 
Supports 
in  inches. 

Breadth 
in  inches. 

Depth 
in  inches. 

Breaking 
Weight 
in  tons. 

Coef- 
ficient of 
Bending 
Strength. 

Remarks. 

Yellow  tine    ...... 

120 

14 

15 

38.15 

2.34 

V 
I2q 

14 

15 

34 

2.09 

>i            < 

AC 

c 

7 

c  .q 

1.62 

0                             I 

4.: 

5 

7 

5.7 

I  .57 

Old  timber 

.(                            < 

4? 

5 

5 

3.  1 

1.6; 

«                             i 

AC 

c 

5 

a.O^ 

1.64 

*i                             < 

1C 

a| 

3i 

.025 

2.O4 

(3ld  timber 

.1                             i 

AC 

2i 

oi 

I   -07^ 

2.  i7 

Pitch  pine             •  .  . 

120 

J4 

IE 

CQ.2=; 

3.64 

«  < 

I2Q 

14. 

1C 

60.2^ 

3.7 

,, 

4; 

5 

7 

7.8 

2.  14 

,« 

AS. 

c 

7 

Q.7C 

2.68 

« 

4D 
4C 

e 

7 

jo.Ct; 

2.  02 

« 

46 

5 

7 

ii 

3.O3 

(, 

4e 

2* 

•U 

i  6 

q    c2 

|, 

AC. 

2* 

^i 

i  .  35 

2.Q7 

AS. 

C 

7 

7 

I  .  QI 

AS. 

c 

7 

8.«; 

2    71 

,,                            « 

AS. 

2i 

•u 

I    125 

2.48 

.<                           (f 

AS. 

2^ 

ol 

I    2 

o  6  i 

American  elm        .  . 

AS. 

7 

MQ 

j.   i 

Old  timber 

AS. 

e 

7 

IS    6 

i    2Q 

K         tt 

«           « 

AC 

ol 

•1 

25c 

.     e    8.1 

0                     «< 

«           t  < 

AS. 

2i 

•7! 

2    6 

57  •} 

«                   X 

Greenheart      

AS. 

C 

7    56 

AS. 

2A 

7 

1  1  .4.5 

6.^1 

, 

AS. 

•i  8^ 

952C 

, 

AS. 

24 

^4 

400 

8  Si 

.4   

R  jiJ  pine           .  .  . 

45 
139 

147 

»i 

9 
6 

3i 
8 

12 

3-55 
24-5 

7    5 

7.82 
8.87 

I    OI 

147 

6 

12 

8.45 

2    15 

A^.5. — The  results  contained  in  the  last  two  tables  are  mainly  deduced  from  experi- 
ments carried  out  under  the  supervision  of  W.  Le  Mesurier,  M.Inst.C.E.,  Dock  Yard,  Liver- 
pool. 


214  THEORY   OF  STRUCTURES. 

THE   WEIGHTS   AND   CRUSHING   WEIGHTS   OF    ROCKS,    ETC. 


Material. 

Weight  per  cu.  ft. 
in  Ibs. 

Crushing-  Weight  in 
Ibs.  per  sq.  in. 

Asphalt  •    

«6 

S  300 

184 

17  2OO 

"         Greenstone*                           ...» 

181 

1  6  SOO 

"         Welsh  

172 

800  to  1,400 

Brick    common                        ......      . 

550  to  800 

2  2^O 

"       Sydney    N    S 

2  2OO 

"       yellow-faced  (Eng  )        ......    . 

ioo  to  135 

I  44O 

"       Staffordshire  blue     

7  2OO 

fire  

I,7OO 

"       pressed  (best)                 .        

jen 

IO  2OO 

112 

Cement    Portland                .            .... 

86  to  04 

i  700  to  6  ooo 

"        Roman              ........ 

IOO 

Clav   

I  IQ 

IIQ 

460  to  775 

"         in  cement       ...      .        .    .  >  .  .  . 

I  "\1 

Earth   

77  to  12^ 

112 

19,600 

Freestone                             

3  ooo  to  3  500 

Glass    flint  

IQ2 

27  5OO 

ic,7 

31,000 

"       common  green        .  . 

1*8 

31  ooo 

172 

16-3 

10  800 

'                 "         red  

16* 

'         Cornish  ,  .  . 

1  66 

14  ooo 

'         Sorrel    

167 

12  8OO 

«         Irish  

10  450 

U    S   (Quincy)    

1  5  ooo 

'         Argyll       

10  900 

06  to  175 

19,600 

Limestone                 .      .  .          

154  to  162 

7  500  to  9  ooo 

Lime    quick.  ...                      

c  a 

86  to  119 

1  20  to  240 

"     (average) 

1  06 

Masonry    common  brick  .  .  .  ) 

500  to  800 

Il6  to  14.4 

760 

"           rubble                           ) 

_4_.  of  cut  stone 

Marble    statuary        ...      ..      

I7O 

3  200 

168  to  170 

8,000  to  9  700 

Oolite    Portland  stone                .            .  .  .  . 

T  C  T 

4  ioo 

"         Bath  stone  

I21? 

177 

"       river                          .  . 

117 

"       pit       

IOO 

QS, 

Sandstone    red  (Eng  )                    

T  O-J 

57OO 

ICQ 

3  ioo 

jc6  to  157 

5  ,  700  to  6  ooo 

"            Scotch     

TC  q  to  155 

5  300  to  7  800 

U.  S  

Sqoo 

Shingle                .... 

83 

Slate    Anglesea   

1  7Q  } 

IO  OOO 

IC7  I 

to 

"      Welsh  

180  1 

24  OOO 

Trao  . 

I7O 

TABLES. 


21$ 


ACTORS    OF    SAFETY. 

GOOD  ORDINARY  WORK. 

Timber 4  to  5  for  dead  load,         8  to  10  for  live  load. 

Metals 3  "  6  "      "       " 

Masonry 4  "        "       "  8  "      "       " 


EXPANSIONS   OF   SOLIDS. 


Materials. 

Linear  Expansion  per  Unit  of  Length. 

Expansion 
in  Bulk. 

From  32°  F.  to 

212°   F. 

From  32°  F.  to 
572°  F. 

From  32°  F. 
to  212°  F. 

Brass  •  

.001868  =  ^s 
.00182      —  ^T 
.001075   =   *>h 

.001718  =  ifa 
.00352    =  ^VT 
.00861    =  TTVr 
.001466  =  ifo 
.00181    =  Ti, 
.00144    =  ^T 
.0002848=  sk 
.000746  =  ^Vtf 
.000884  =  TT-VT 
.001909  =  3fo 
.001079  =  ^7 

.00124      =  B^T 

.002173  =  -flv 
.001235  =  gl7 

.001182  =  g-^ 
.002941  =  ^0 

.003108    =   si  3 

.001883  =  ^ 
.001468  =  ^T 

.0065 
.0054 
.0033 
.0055 

.0027 
.0057 

.0036 
.0066 

.0036 

.0058 

Bronze      . 

Fir      

Glass  .  .                            

Gold  

I  ron  wire            

Lead           

Oak                .                .... 

Silver  

Steel,  unhardened  

"      hardened  .  .  . 

Tin   .  . 

\Vrought-iron  (bar)             . 

"         (tor  smith-work).. 
Zinc,  cast    

"     hammered  

2l6  THEORY  OF  STRUCTURES. 


EXAMPLES. 

1.  How  many  square  inches  are  there  in  the  cross-section  of  an  iron 
rail  weighing  30  Ibs.  per  lineal  yard?     How  many  in  a  yellow-pin.e  beam 
of  the  same  lineal  weight?  Ans.  3  sq.  in.;  45  sq.  in. 

2.  A  vertical  wrought-iron  bar  60  ft.  long  and  I   in.  in  diameter  is 
fixed  at  the  upper  end  and  carries  a  weight  of  2000  Ibs.  at  the  lower  end. 
Find  the  factors  of  safety  for  both  ends,  the  ultimate  strength  of  the 
iron  being  50,000  Ibs.  per  sq.  in.  Ans.   i9T9f  ;  i8T372T. 

3.  A  vertical  rod  fixed  at  both  ends  is  weighted  with  a  load  w  at  an 
intermediate  point.     How  is  the  load  distributed  in  the  tension  of  the 
upper  and  compression  of  the  lower  portion  of  the  rod? 

Ans.  Inversely  as  the  lengths. 

4.  'Find  the  length  of  a  steel  bar  of  sp.  gr.  7.8  which,  when  suspended 
vertically,  would  break  by  its  own  weight,  the  ultimate  strength  of  the 
metal  being  60,000  Ibs.  per  sq.  in.  Ans.  17,723  ft. 

5.  The  iron  composing  the  links  of  a  chain  is  %  in.  in  diameter;  the 
chain  is  broken  under  a  pull  of  10,000  Ibs.     What  is  the  corresponding 
tenacity  per  sq.  in.  ?  Ans.  57,272^  Ibs. 

6.  A  vertical  iron  suspension-rod  90  ft.  long  carries  a  load  of  20,000 
Ibs.  at  its  lower  end  ;  the  rod  is  made  up  of  three  equal  lengths  square 
in  section.     Find  the  sectional  area  of  each  length,  the  ultimate  tenacity 
of  the  iron  being  50,000  Ibs.  per  sq.  in.,  and  5  a  factor  of  safety. 


7.  If  the  rod  in  the  previous  question  is  of  a  conical  form,  what 
should  be  the  area  of  the  upper  end  ?     Also  find  the  intensities  of  the 
tension  at  30  and  60  ft.  from  the  lower  end. 

Ans.  2.0407  sq.  in.;  9999.612  Ibs.,  9999.605  Ibs.  per  sq.  in. 

8.  The  dead  load  of  a  bridge  is  5  tons  and  the  live  load   10  tons  per 
panel,  the  corresponding  factors  of  safety  being  3  and  6.     If  the  two  loads 
are  taken  together,  making  9  tons  per  panel,  what  factor  of  safety  would 
you  use  ?  Ans.  5. 

9.  The  end  of  a  beam  10  in.  broad  rests  on  a  wall  of  masonry.     If  it  be 
loaded  with  10  tons,  what  length  of  bearing  surface  is  necessary,  the  safe 
crushing  stress  for  stone  being  150  Ibs.  per  sq.  in.  ?  Ans.  13^-  in. 

10.  Find  diameter  of  bearing  surface  at  the  base  of  a  column  loaded 
with  20  tons,  the  same  stress  being  allowed  as  in  the  preceding  question. 

Ans.    \  '380.  1  2. 


EXAMPLES.  217 

1 1 .  In  the  chain  of  a  suspension-bridge  five  flat  links  dovetail  with  four 
alternately,  and  a  cylindrical  pin  passes  through  the  eyes.     The  pull  on 
the  chain  is  200  tons.     Find  the  area  of  the  pin,  the  bearing  strength  of 
the  metal  being  6  tons  per  sq.  in.  Ans.  if  sq.  in. 

12.  An  iron  bar  of   uniform  section  and   10  ft.  in   length  stretches 
.12  in.  under  a  unit  stress  of  25,000  Ibs.     Find  E. 

Ans.  25,000,000  Ibs. 

13.  A  ship  at  the  end  of  a  6oo-ft.  cable  and  one  at  the  end  of  a  5oo-ft. 
cable  stretch  the  cables  3  in.  and  2$  in.,  respectively.     What  are  the  cor- 
responding strains?  Ans.  y^ff. 

14.  A  rectangular  timber  tie  is  12  in.  deep  and  40  ft.  long.     If  E  = 
1,200,000  Ibs.,  find  the  proper  thickness  of  the  tie  so  that  its  elongation 
under  a  pull  of  270,000  Ibs.  may  not  exceed  1.2  in.  Ans.  J\  in. 

15.  A  wrought-iron  bar  60  ft.  long  is  stretched  5  in.  by  a  pull  of 
5000  Ibs.     Find  its  diameter,  E  being  25,000,000  Ibs.  Ans.  .59  in. 

1 6.  A  wrought-iron  rod  984  ft.  long  alternately  exerts  a  thrust  and  a 
pull  of  52,910  Ibs. ;  its  cross-section  is  9.3  sq.  in.     Find  the  loss  of  stroke, 
E  being  29,000,000  Ibs.  Ans.  4.632  in. 

17.  A  wrought-iron  bar  2  sq.  in.  in  sectional  area  has  its  ends  fixed 
between  two  immovable  blocks  when  the  temperature  is  at  32°  F.     If 
E  =  29,000,000  Ibs.,  what  pressure  will  be  exerted  upon  the  blocks  when 
the  temperature  is  100°  F.  ?  Ans.  27388!  Ibs. 

1 8.  What  should  be  the  diameter  of  the  stays  of  a  boiler  in  which  the 
pressure  is  30  Ibs.  per  sq.  in.,  allowing  one  stay  to  each   i£  sq.  ft.  of 
surface  and  a  stress  of  3500  Ibs.  per  sq.  in.  of  section  of  iron  ? 

Ans.   1 1  in. 

19.  A  force  of  10  Ibs.  stretches  a  spiral  spring  2  in.     Find  the  work 
done  in  stretching  it  successively  i  in.,  2  in.,  3  in.,  up  to  6  in. 

Ans.  f,  -V-,  -V-,  -8/->  -!•£*,  ip-  in. -Ibs. 

20.  A  roof  tie-rod  142  ft.  in  length  and  4  sq.  in.  in  sectional  area  is 
subjected   to  a  stress  of  80,000   Ibs.     If  E  =  30,000,000  Ibs.,  find  the 
elongation  of  the  rod  and  the  corresponding  work. 

Ans.   1. 136  in.;  3786!  ft.-lbs. 

21.  An  iron  wire  i  in.  in  diameter  and  250  ft.  in  length  is  subjected 
to  a  tension  of  600  Ibs.,  the  consequent  strain  being  ^^.     Find  E,  and 
show  by  a  diagram  the  amount  of  work  done  in  stretching  the  wire 
within  the  limits  of  elasticity.  Ans.   14,661,818^  Ibs. 

22.  A  timber  pillar  30  ft.  in  length  has  to  support  a  beam  at  a  point 
30  ft.  from  the  ground.     If  the  greatest  safe  strain  of  the  timber  is  3-^, 
what  thickness  of  wedge  should  be  driven  between  the  head  of  the 
pillar  and  the  beam?  Ans.  T^  ft. 


2l8  THEORY   OF   STRUCTURES. 

23.  An  hydraulic  hoist-rod  50  ft.  in  length  and   i   in.  in  diameter  is 
attached  to  a  plunger  4  in.   in   diameter,  upon  which  the   pressure   is 
800  Ibs.  per  sq.  in.     Determine  the  altered  length  of  the  rod,  E  being 
30,000,000  Ibs.  Ans.  .0213  ft. 

24.  A  short  cast-iron  post  is  to  sustain  a  thrust  of    icoo  Ibs.,  the  ul- 
timate crushing  strength  of  the  iron  being  80,000  Ibs.  per  sq.  in.  and  10 
a  factor  of  safety.     Find  the  dimensions  of  the  post,  which  is  rectangular 
in  section  with  the  sides  in  the  ratio  of  2  to  i.  Ans.  4  in.;  2  in. 

25.  The  length  of  a  cast-iron  pillar  is  diminished  from  20  ft.  to  19.97 
ft.  under  a  given  load.     Find  the  strain  and  the  compressive  unit  stress, 
E  being  17,000,000  Ibs.  Ans.  .0015  ;  25,500  Ibs.  per  sq.  in. 

26.  A  rectangular  timber  strut  24  sq.  in.  in  sectional  area  and  6  ft.  in 
length   is  subjected   to   a   compression    of    14,400   Ibs.     Determine   the 
diminution  of  the  length,  E  being  1,200,000  Ibs.  Ans.   .003  ft. 

27.  Find  the  height  from  which  a  weight  of  200  Ibs.  may  be  dropped 
so  that  the  maximum  admissible  stress  produced  in  a  bar  of  i  sq.  in. 
section  and   5   ft.  long  may  not  exceed  20,000  Ibs.  per  sq.  in.,  the  co- 
efficient of  elasticity  being  27,000,000  Ibs. 

Ans.  -fa  ft.,  or,  more  accurately,  -/-fa  ft. 

28.  Find  the  H.  P.  required  to  raise  a  weight  of  10  tons  up  a  grade 
of  i  in  12  at  a  speed  of  6  miles  per  hour  against  a  resistance  of  9  Ibs.  per 
ton.  Ans.  31.3. 

29.  A  square  steel  bar  10  ft.  long  has  one  end  fixed  ;  a  sudden  pull  of 
40,000  Ibs.  is  exerted  at  the  other  end.     Find   the  sectional  area  of  the 
bar  consistent  with  the  condition  that  the  strain  is  not  to  exceed  yl^. 
E  =  30,000,000  Ibs.     Find  the  resilience  of  the  bar. 

Ans.  2  sq.  in. ;  533^  ft.-lbs. 

30.  How  much  work  is  done  in  subjecting  a  cube  of  125  cu.  in.  of 
iron  to  a  tensile  stress  of  3000  Ibs.  per  sq.  in.  ?  Ans.   1 1^  ft.-lbs. 

31.  A  signal-wire  2000  ft.  in  length  and  -J-  in.  in  diameter  is  subjected 
to  a  steady  stress  of  300  Ibs.     The  lever  is  suddenly  pulled  back,  and  the 
corresponding  end  of  the  wire  moves  through  a  distance  of  4  in.     De- 
termine the  instantaneous  increase  of  stress.  Ans.  51  Iff  Ibs. 

32.  If  the  total  back-weight  is  350  Ibs.,  what  is  the  range  of  the  sig- 
nal end  of  the  wire  ?  Ans.  T^f^  ft. 

33.  A  steel  rod  of  length  L  and  sectional  area  A  has  its  upper  end 
fixed  and  hangs  vertically.     The  rod  is  tested  by  means  of  a  ring  weigh- 
ing 60  Ibs.  which  slides  along  the  rod  and  is  checked  by  a  collar  screwed 
to  the  lower  end.     A  scale  is  marked  upon  the  rod  with  the  zero  at  the 
fixed  end.     If  the  strain  in  the  steel  is  not  to  exceed  T^-Gt  what  is  the 
reading  from  which  the  weight  is  to  be  dropped  ?     What  should  be  the 
reading  of  the  collar?    E  =  35,000,000  Ibs. 

Ans.  Distance  from  point  of  suspension  =  (f§J  —  \\A)L ;  y 


EXAMPLES.  219 

34.  A  load  of  looolbs.  falls  i  in.  before  commencing  to  stretch  a  sus- 
pending rod  by  which  it  is  carried.     If  the  sectional  area  of  the  rod  is 
2  sq.  in.,  length  100  in.,  and  E  =  30,000,000  Ibs.,  find  the  stress  pro- 
duced. Ans.   17,828  Ibs.  per  sq.  in. 

35.  If  the  rod  carries  a  load  of  5000  Ibs.,  and  an  additional  load  of 
2000  Ibs.  is  suddenly  applied,  what  is  the  stress  produced  ? 

Ans.  4500  Ibs.  per  sq.  in. 

36.  Steam  at  a  pressure  of  50  Ibs.  per  sq.  in.  is  suddenly  admitted 
upon    a    piston  32  in.  in  diameter.     The  steel    piston-rod  is  48  in.  in 
length  and  2  in.  in  diameter,  E  being  35,000,000  Ibs.     Find  the  work 
done  upon  the  rod.  Ans.  117.69  ft. -Ibs. 

37.  What  should  be  the  pressure  of  admission  to  strain  the  rod  to  a 
proof  of  .001  ?  Ans.  68ff  Ibs.  per  sq.  in. 

38.  A  boulder-grappler  is  raised  and  lowered  by  a  wire  rope  i  in.  in 
diameter  hanging  in  double  sheaves.     On  one  occasion  a  length  of  150 
ft.  of  rope  was  in  operation,  the  distance  from  the  winch  to  the  upper 
block  being  30  ft.     The  grappler  laid  hold  of  a  boulder  weighing  20,000 
Ibs.     What  was  the  extension  of  the  rope,  E  being  15,000,000  Ibs.  ? 

Ans.  Tjo  ft. 

39.  The  boulder  suddenly  slipped  and  fell  a  distance  of  6  in.  before 
it  was  again  held.     Find  the  maximum  stress  upon  the  rope. 

Ans.  50,452-^  Ibs.  per  sq.  in. 

40.  What  weight  of  boulder  may  be  lifted  if  the  proof-stress  in  the 
rope  is  not  to  exceed  25,000  Ibs.  per  sq.  in.  of  gross  sectional  area? 

Ans.  78.571!  Ibs. 

41.  The  steady  thrust  or  pull  upon  a  prismatic  bar  is  suddenly  re- 
versed.    Show  that  its  effect  is  trebled. 

42.  A  weight  W  is  suspended  by  a  spring,  which  it  stretches.     The 
weight  is  further  depressed  i  ft.,  when  it  is  suddenly  released  and  allowed 
to  oscillate.      Find   its  velocity  at  a  distance  x  from   the  position   of 
equilibrium.  /  ~ 

Ans.  y  10(1  -  icur2)-^. 

43.  If  a  spring  deflects  .001  ft.  under  a  load  of  i  lb.,  what  will  be  the 
period  of  oscillation  of  a  weight  of  14  Ibs.  upon  the  spring? 

44.  Show  that  the  change  of  a  unit  of  volume  of  a  solid  body  under 

a  longitudinal  stress  is  Af  i 1,  which  becomes  —  if  m  =  4,  as  in  metals, 

and  nil  when  m  =  2,  as  in  india-rubber  (page  142). 

45.  A  steel  bar  stretches  YgVffth  of  its  original  length  under  a  stress  of 
20,oco  Ibs.  per  sq.  in.     Find  the  change  of  volume  and  the  work  done 
per  cubic  inch.  Ans.  ^V^th  ;  f  ft.-lb.  per  cu.  in. 


220  THEORY  OF  STRUCTURES. 

46.  During  the  plastic  deformation  of  a  prismatic  bar,  show  that  the 
change  in  sectional  area  is  proportional  to  the  deformation  calculated  on 
the  altered  length  of  the  bar. 

47.  A  prismatic  bar  of  volume  V  changes  in  length  from  L  to  L  ±  x 
.inder  the  "  fluid  pressure"/.     Find  the  corresponding  work. 

Ans.  /  Flog.CZ  ±  .r). 

48.  Show  that  the  total  work  done  in  raising  a  number  of  weights 
through  to  a  given  level  is  the  product  of  the  sum  of  the  weights  and  the 
vertical  displacement  of  their  centre  of  gravity. 

49.  An  engine  has  to  raise  4000  Ibs.  icooft.  in  5  minutes.    What  is  its 
H.  P.  ?     How  long  will  the  engine  take  to  raise  10,000  Ibs.  100  ft.  ? 

Ans.  24/3-  H.  P.  ;  i^  min. 

50.  How  many  men  will  do  the  same  work  as  the  engine  in  the  pre- 
ceding question,  assuming  that  a  man  can  do  900,000  ft.  -Ibs.  of  work  in 
a  day  of  9  hours  ?  Ans.  480  men. 

51.  Determine  the  H.  P.  which  will  be  required  to  drag  a  heavy  rock 
weighing  10  tons  at  the  rate  of  10  miles  an  hour  on  a  level  road,  the 
coefficient  of  friction  being  0.8.     What  will  be  the  speed  up  a  gradient 
of  i  in  50,  the  same  power  being  exerted  ? 

Ans.  477|f  ;  9f|  miles  per  hour. 

52.  Two  horses  draw  a  load  of  4000  Ibs.  up  an  incline  of  i  in  25  and 
1000  ft.  long.     Determine  the  work  done.  Ans.   160,000  ft.-lbs. 

53.  At  what  speed  do  the  horses  walk  if  each  horse'  does  16,000  ft.- 
lbs.  of  work  per  minute  ?  Ans.  2T3T  miles  per  hour. 

54.  A  wrought-iron  rod  25  ft.  in  length  and  i   sq.   in.   in  sectional 
area  is  subjected  to  a  steady  stress  of  5000  Ibs.     What  amount  of  live 
load  will  instantaneously  elongate  the  rod  by  i  in.,  E  being  30,000,000 
Ibs.  ?  Ans.  6250  Ibs. 

55.  Determine  the  shortest  length  of  a  metal  bar  a  sq.  in.  in  sec- 
tional area  that  will  safely  resist  the  shock  of  a  weight  of  W  Ibs.  falling 
n.  distance  of  h  ft.     Apply  the  result  to  the  case  of  a  steel  bar  i  sq.  in.  in 
sectional  area,  the  weight  being  50  Ibs.,  the  distance  16  ft.,  the  proof- 
strain  Tj-^-g-,  and  E  =  35,000,000  Ibs. 

lEWah 
Ans.    —^—^  ----  ,/  being  the  safe  unit  stress  ;  ^H^  ft. 


, 

56.  A  shock  of  N  ft.-lbs.  is  safely  borne  by  a  bar  /  ft.  in  length  and  a 
sq.  in.  in  sectional  area.     Determine  the  increased  shock  which  the  bar 
will  bear  when  the  sectional  area  of  the  last  wth  of  its  length  is  increased 

to  ra.  f         i          i  \ 

Ans.  N(i  ---  1  --  . 
\         m       rm] 

57.  The  bar  in  Example  12  is  i  sq.  in.  in  section.     Determine  the  work 
stored  up  in  the  rod  in  foot-pounds  and  compare  it  with  the  work  which 


EXAMPLES.  221 

would  be  stored  up  if  for  half  its  length  the  rod  has  its  section  increased 
to  4  in.  Ans.    125  ft.-lbs.;  f  of  125  ft.-lbs. 

58.  If  25,000  Ibs.  per  sq.  in.  is  the  proof-stress,  find  the  modulus  of 
resilience  for  the  i-in.  rod.  Ans.   25  in  in.-lb.  units. 

59.  A  steel  rod  100  ft.  in  length  has  to  bear  a  weight  of  4000  Ibs.     If 
E  —  35,000,000  Ibs.,  and  if  the  safe  strain  is  .0005,  determine  the  sectional 
area  of  the  rod  (i)  when  the  weight  of  the  rod   is  neglected  ;  (2)  when 
the  weight  of  the  rod  is  taken  into  account.     Also  in  the  former  case, 
determine  the  work  done  in  stretching  the  rod  -fa  in.,  -£$  in.,  y3^  in.,  .  .  . 
T6¥  in.,  successively. 

Ans.  &  sq.  in.  ;  -fflfo  sq.  in.  ;  33$,  133^,  300,  ...  1200  in.  -Ibs. 
60    A  line  of  rails  is  10  miles  in  length  when  the  temperature  is  at. 
32°  F.     Determine  the  length  when  the  temperature  is  at  100°  F..  and 
the  work  stored  up  in  the  rails,  E  being  30,000,000  Ibs. 

Ans.   10.008  miles  ;  10.24  H.  P. 

61.  A  wrought-  iron  bar  25  ft.  in  length  and  i  sq.  in.  in  sectional  area 
stretches  .0001745  ft.  for  each  increase  of  i°  F.  in  the  temperature.     If 
E  =  29,000,000  Ibs.,  determine  the  work  done  by  an  increase  of  20°  F. 

How  may  this  property  of  extension  under  heat  be  utilized  in  straight- 
ening walls  that  have  fallen  out  of  plumb  ?  Ans.  7.064  ft.-lbs. 

62.  Find  the  work  done  in  raising  a  Venetian  blind,  w  being  the 
weight  of  a  slat,  a  the  distance  between  consecutive  slats,  and   n  the 

number  of  slats.  n(n  +  i) 

Ans.    wa  ----  . 

2 

63.  How  many  |-in.  rivets  must  be  used  to  join  two  wrought-iron 
plates,  each  36  in.  wide  and  £  in.  thick,  so  that  the  rivets  may  be  as 
strong  as   the   riveted   plates,   the    tensile    and   shearing   strength   of 
wrought-iron  being  in  the  ratio  of  10  to  9?  Ans.  17  rivets  (16.3). 

64.  A  horizontal  string,  without  weight,  of  length  20,  and  sectional 
area  S,  has  its  two  ends  fixed  in  the  same  horizontal  plane.     A  weight 

W  suspended  from  its  centre  draws  the  string  slightly  out  of  the  hori- 
zontal.    Show  that,  approximately, 


t  being  the  intensity  of  the  tension,  d  the  depression,  and  E  the  coef- 
ficient of  elasticity. 

65.  A  heavy  wire  of  length  2a,  sectional  area  S,  and  weight  IV  has  its 
ends  fixed  in  a  horizontal  plane  and  is  allowed  to  deflect  under  its  own 
weight.  Find  the  deflection  d  and  the  tenacity  /  (assumed  uniform 
throughout). 


222  THEORY  OF  STRUCTURES. 

66.  A  length  270  ft.  of  wire  i  sq.  in.  in  section  and  of  sp.  gr.  7.8  is 
subjected  to  the  above  conditions.     Find  the  tenacity  of  the  wire  and  the 
deflection,  the  coefficient  of  elasticity,  E,  being  25.300,000  Ibs. 

67.  A  brick  wall  2  ft.  thick,   12  ft.  high,  and  weighing  112  Ibs.  per 
cu.  ft.    is  supported  upon  solid  pitch-pine  columns  9  in.  in  diameter, 
10  ft.  in  length,  and  spaced  12  ft.  centre  to  centre.     Find  the  compress- 
ive  unit  stress  in  the  columns  (i)  at  the  head;  (2)  at  the  base.    The  tim- 
ber weighs  50  Ibs.  per  cu.  ft.  Ans.  507.03  Ibs. ;  510.5  Ibs. 

68.  If  the  crushing  stress  of  pitch-pine  is  5300  Ibs.  per  sq.  in.  and  the 
factor  of  safety  10,  find  the  height  to  which  the  wall  may  be  built. 

Ans.  12.46  ft. 

69.  Determine   the   diameter   of   the   wrought-iron   columns   which 
might  be  substituted  for  the  timber  columns  in  question  67,  allowing  a 
working  stress  in  the  metal  of  7500  Ibs.  per  sq.  in.  Ans.   2.36  in. 

70.  Find  the  greatest  length  of  an  iron  suspension-rod  which  will  carry 
its  own  weight,  the  stress  being  limited  to  4  tons  per  sq.  in.     What  will 
be  the  extension  under  this  load,  E  being  12,500  tons? 

Ans.  2700  ft. ;  .864  ft. 

71.  A  horizontal  cast-iron  bar  i  ft.  long  exactly  fits  between  two  verti- 
cal plates  of  iron.     How  much  should  its  temperature  be  raised  so  that 
it  might  remain  supported  between  the  plates  by  the  friction,  the  coef- 
ficient of  friction  being  -\  ?  Ans,  ^°  F. 

72  The  fly-wheel  of  a  40  H.  P.  engine,  making  50  revolutions  per 
minute,  is  20  ft.  in  diameter  and  weighs  12,000  Ibs.  What  is  its  kinetic 
energy? 

If  the  wheel  gives  out  work  equivalent  to  that  done  in  raising  5000 
Ibs.  through  a  height  of  4  ft.,  how  much  velocity  does  it  lose? 

The  axle  of  the  fly-wheel  is  12  in.  in  diameter.  What  proportion  of 
the  H.  P.  is  required  to  turn  the  wheel,  the  coefficient  of  friction  being 
.08? 

If  the  fly-wheel  is  disconnected  from  the  engine  when  it  is  making 
50  revolutions  per  minute,  how  many  revolutions  will  it  make  before  it 
comes  to  rest  ? 

Ans.  511,260.4  ft. -Ibs.  ;  1.04  ft.  per  sec. ;  |ths  ;  169.4. 

73.  The  velocity  of  flow  of  water  in  service-pipe  48  ft.  long  is  64  ft. 
per  sec.     If  the  stop-valve  is  closed  in  -^  of  a  sec.,  find  the  increase  of 
pressure  near  the  valve.  Ans.  375  Ibs.  per  sq.  in. 

74.  Work  equivalent  to  50  ft.-lbs.   is  done  upon  a  bar  of  constant 
sectional  area,  and  produces  in  it  a  uniform  tensile  stress  of  10,000  Ibs. 
per 'sq.  in.     Find  the  cubic  content  of  the  bar,  E  being  30,000,000. 

Ans.  360  cu.  in. 

75.  A  fly-wheel  weighs  20  tons  and  its  radius  of  gyration  is  5  ft.    How 


EXAMPLES.  .  223 

much  work  is  given  out  while  the  speed  falls  from  60  to  50  revolutions 
per  minute  ?  Ans.  94TT36T  ft.-tons, 

76.  The  resilience  of  an  iron  bar  i  sq.  in.  in  section  and  20  ft.  long  is 
30,000  ft.-lbs.     What  would  be  the  resilience  if  for  19  ft.  of  its  length  it 
was  composed  of  iron  2  sq.  in.  in  section,  the  remaining  foot  being  the 
same  size  as  before?  Ans.  8625  ft.-lbs. 

77.  A  particle  under  the  action  of  a  number  of  forces  moves  with  a 
uniform  velocity  in  a  straight  line.     What  condition   must  the  forces 
fulfil  ?  Ans.   Equilibrium. 

78.  Determine  the  constant  effort  exerted  by  a  horse  which   does 
1,650,000  ft.-lbs.  of  work   in  one  hour  when  walking  at  the  rate  of  2$ 
miles  per  hour.  Ans.   125  Ibs. 

79.  A  train  is  drawn  by  a  locomotive  of  160  H.  P.  at  the  rate  of  60 
miles  an  hour  against  a  resistance  of  20  Ibs.  per  ton.     What  is  the  gross 
weight  of  the  train  ?  Ans.   50  tons. 

80.  A  train  of  292!  tons  is  drawn  up  an  incline  of  I  in  75,  5^  miles 
long,  against  a  resistance  of  10  Ibs.  per  ton,  in  ten  minutes.     Find  the 
H.  P.  of  the  engine.     The  speed  on  the  level,  the  engine  exerting  769.42 
H.  P.,  is  43.4  miles  per  hour.    What  is  the  resistance  in  pounds  per  ton  ? 

Ans.   1027  H.  P.  ;  22.7  Ibs.  per  ton. 

81.  The  dead  load  upon  a  short  hollow  cast-iron   pillar  with  a  sec- 
tional area  of  20  sq.  in.  is  50  tons  (of  2000  Ibs.).     If  the  strain  in  the 
metal  is  not  to  exceed  .0015,  find   the  greatest  live  load  to  which  the 
pillar  might  be  subjected,  E  being  17,000,000  Ibs.         Ans.   255,000  Ibs. 

82.  A  steel  suspension-rod  30  ft.  in  length  and  \  sq.  in.  in  sectional 
area  carries  3500  Ibs.  of  the  roadway  and  3000  Ibs.  of  the  live  load.     De- 
termine the  gross  load   and  also   the  extension  of   the  rod,  E  being 
35,000.000  Ibs.  Ans.  yf^  ft. 

83.  A  steel  rod    10  ft.  in  length   and  \  sq.  in.  in   sectional  area  is 
strained  to  the  proof  by  a  tension  of  25,000  Ibs.     Find  the  resilience  of 
the  rod,  E  being  35,000,000  Ibs.  Ans.   178*  ft.-lbs, 

84.  What  form  does  the  useful  work  done  by  a  hammer  take  when  a 
nail  is  driven  into  any  material  ?    What  becomes  of  the  rest  of  the  energy 
of  the  mass  of  the  hammer  after  striking  the  blow? 

85.  A  hammer  weighing  2  Ibs.  strikes  a  steel  plate  with  a  velocity  of 
10  ft.  per  sec.,  and  is  brought  to  rest  in  .0001  sec.     What  is  the  average 
force  on  the  steel  ?  Ans.  6250  Ibs. 

86.  A  hammer  weighing  lolbs.  strikes  a  blow  of  10  ft.-lbs.  and  drives 
a  nail  5  in.  into  a  piece  of  timber.     Find  the  velocity  of  the  hammer  at 
the  moment  of  contact,  and"  the  mean  resistance  to  entry.     Also  find  the 
steady  pressure  that  will  produce  the  same  effect  as  the  hammer. 

Ans.  8  ft.  per  sec. ;  240  Ibs. ;  480  Ibs. 


224  THEORY   OF  STRUCTURES. 

87.  When  a  nail  is  driven  into  wood,  why  do  the  blows  seem  to  have 
little  if  any  effect  unless  the  wood  is  backed  up  by  a  piece  of  metal  or 
stone  ? 

88.  In  Question  86,  taking  the  weight  of  the  nail  to  be  4  oz.  and  the 
weight  of  the  piece  of  timber  to  be  100  Ibs.,  find  the  depth  and  time  of 
the  penetration  (a)  when  the  timber  is  fixed ;  (b)  when  the  timber  is  free 
to  move. 

Also  in  case  (b)  find  the  distance  through  which  the  timber  moves. 
Ans. — (a)  ff  in. ;  -fa  sec. 

(b)  .44245  in.;  .0009448  sec.  ;  .04113  in. 

\  89.  Show  that  the  greater  part  of  the  energy  of  impact  is  expended 
in  local  damage  at  high  velocities,  and  in  straining  the  impinging  bodies 
as  a  whole  at  low  velocities. 

90.  A  pile-driver  of  300  Ibs.  falls  20  ft.,  and  is  stopped  in  -fa  sec.   What 
is  the  average  force  exerted  on  the  pile  ?  Ans.  3344  Ibs. 

91.  A  weight  falls  16  ft.  and  does  2560  ft.-lbs.  of  work  upon  a  pile 
which   it  drives  4  in.  against  a  uniform  resistance.     Find  the  weight  of 
the  ram,  and  the  resistance.  Ans.  160  Ibs. ;  7680  Ibs. 

92.  A  pitch-pine  pile  14  in.  square  is  20  ft.  above  ground,  and  is 
being  driven  by  a  falling  weight  of  112  Ibs.     It  E  =  1,500,000  Ibs.,  find 
the  fall  so  that  the  inch-stress  at  the  head  of  the  pile  may  be  less  than 
800  Ibs. 

Supposing  that  the  pile  sinks  2  in.  into  the  ground,  by  how  much 
would  it  be  safe  to  increase  the  fall? 

Ans.  7.456  ft. ;  116.5  ft- 

93.  A  weight  of  W\  tons  falls  h  ft.,  and  by  n  successive  instantaneous 
blows  drives  an  inelastic  pile  weighing  W*  tons  a  ft.  into  the  ground. 
Assuming  the  pile  and  weight  to  be  inelastic,  find  (a)  the  mean  effective 
resistance  of  the  ground. 

If  the  ground-resistance  increases  directly  as  the  depth  of  penetration, 
find  (b)  how  far  the  pile  will  sink  nnder  the  rth  blow.  If  the  head  of  the 
pile  is  crushed  for  a  length  of  x  ft.,  x  being  very  small  as  compared  with 

CL 

the  depth  —  of  penetration,  find  (i)  the  mean  thrust,  during  the  blow, 
n 

between  the  weight  and  hammer ;  (2)  the  time  of  penetrating  the  ground  ; 
(3)  the  time  during  which  the  blow  acts. 

W^      nh  a 

l^/r1  -|.  w^  a  >  n 

(0    ~j*7~T~iT7  T--      (2)  TTr 


94.   An   inelastic   pile  weighing  788  Ibs.  is  driven   3^  feet  into  tile 
ground  by  120  blows  from  a  weight  of  112  Ibs.  falling  30  ft.     Find  the 


EXAMPLES.  22$ 

steady  load  upon  the  pile  which  will  produce  the  same  effect,  assuming 
the  ground-resistance  to  be  (a)  uniform  ;  (b)  proportional  to  the  depth  of 
penetration.  If  the  resistance  is  uniform,  how  long  (c)  does  each  move- 
ment of  the  pile  last  ?  How  many  blows  (d)  are  required  to  drive  the 
pile  the  first  half  of  the  depth,  viz.,  if  ft.,  the  ground-resistance  being 
7168  Ibs.  ?  How  far  (e)  does  the  pile  sink  under  the  last  blow? 

Ans.    (a)    14,336  Ibs. ;    (b)   28,672  Ibs.  ;    (c)  .0107  sec. ;    (d}  30  ; 
(e)  .016  in. 

95.  A  steamer  of  8000  tons  displacement  sailing  due  east  at  16  knots 
an  hour  collides  with  a  steamer  of  5000  tons  displacement  sailing  at  10 
knots  an  hour.     Find  the  energy  of  collision  if  the  latter  at  the  moment 
of  collision  is  going  (i)  due  west ;  (2)  north-west ;  (3)  north-east. 

96.  A  hammer  weighing  2  Ibs.  strikes  a  nail  with  a  velocity  of  15  ft. 
per  sec.,  driving  it  in  £  in.     What  is  the  mean  pressure  overcome  by  the. 
nail  ?  Ans.  673  Ibs. 

97.  A  beam  will  safely  carry  i  ton  with  a  deflection  of  i  in.     From 
what  height  may  a  weight  of  100  Ibs.  drop  without  injuring  it,  neglecting 
the  effect  of  inertia?  Ans.  11.2  in. 

98.  A  rifle-bullet  45  in.  in  diameter  weighs  i  oz. ;  the  charge  of  pow- 
der weighs  85  grains;  the  muzzle-velocity  is  1350  ft.  per  sec. ;  the  weight. 
of  the  rifle  is  9  Ibs.     Neglecting  the  twist  determine  the  energy  of  i  Ib.  of 
powder.     If  the  bullet  loses  \  of  its  velocity  in  its  passage  through  the 
air,  find  the  average  force  of  the  blow  on  the  target  into  which  the  bullet: 
sinks  i  in. 

If  there  is  a  twist  of  i  in  20  in.,  find  the  charge  to  give  the  same 
muzzle- velocity,  the  length  of  the  barrel  being  33  in. 

99.  A  leather  belt  runs  at  2400  ft.  per  minute.     Find  how  much  its 
tension  is  increased  by  centrifugal  action,  the  weight  of  leather  being 
taken  at  60  Ibs.  per  cubic  foot.  Ans.  2of  Ibs. 

100.  Find  the  centrifugal  force  arising  from  a  cylindrical  crank- pin 
6  in.  long  and  3^  in.  in  diameter,  the  axis  of  the  pin  being  12  in.  from 
the  axis  of  the  engine-shaft,  which  makes  200  revolutions  per  minute. 
How  would  you  balance  such  a  pin  ?  Ans.  55.02  Ibs. 

101.  The  pull  on  one  of  the  tension-bars  of  a  lattice  girder  fluctuates 
from  12.8  tons  to  4  tons.     If  24  tons  is  the  statical  breaking  strength  of 
the  metal,  15  tons  the  primitive  strength,  determine  the  sectional  area, 
of  the  bar,  3  being  a  factor  of  safety.        Ans.  2.15  sq.  in.  (Launhardt)  ; 

1.87  sq.  in.  (Unwin). 

102.  The  stress  in  a  diagonal  of  a  steel  bowstring  girder  fluctuates 
from   a  tension   of   15.15  tons  to  a  compression  of  7.65  tons.     If  the 
primitive  strength  of  the  metal  is  24  tons  and  the  vibration  strength  12: 


226  THEORY  OF  STRUCTURES. 

ions,  find  the  proper  sectional  area  of  the  diagonal,  3  being  a  factor  of 
safety.        Ans.  2.53  sq.  in.  (Weyrauch)  ; 

1.7  sq.  in.   (Unvvin),  40  tons  per  sq.  in.  being  statical 
strength. 

103.  A  wrought-iron  screw-shaft  is  driven  by  a  pair  of  cranks  set  at 
^ight    angles.     Neglecting   the   obliquity  of  the   connecting-rods,  and 
assuming  that  the  pull  on  the  crank-pin  is  constant,  compare  the  coef- 
rficients  of  strength  (a'  and  /)  to  be   used  in  calculating  the  diameter  of 
a. he  shaft.     How  is  the  result  affected  by  the  stopping  of  the  engine  ? 

Ans.  a'  —  .82*;  a'  —  £/. 

104.  Taking/ =  EA.  as  the  ordinary  analytical  expression  of  Hooke's 
Law,  find  the  value  of  the  modulus  of  elasticity  when  calculated  (i)  from 
the  actual  stress  and  the  elongation  per  unit  of  initial  length  ;  (2)  from 
the  actual  stress  and  the  elongation  per  unit  of  stretched  length. 

Ans.   (i)  E+f;  (2)  E+f(i  +  A)2  =  £•+/(!  +  2A),  if  A  is  small. 

105.  In  a  fly-wheel  weighing  12,000  Ibs.  and  making  50  revolutions 
per  minute,  the  centre  of  gravity  is  one  seventeenth  of  an  inch  out  of  the 
centre.     Find  the  centrifugal  force.  Ans.  50.4  Ibs. 

106.  In  the  preceding  question,  if  the  axis  of  rotation  is  inclined  to 
the   plane    of   the  wheel  at   an    angle    cot-'.ooi,  find    the  centrifugal 
couple,  the  radius  of  gyration  being  10  ft.  Ans.   1028.9  ft.-lbs. 

107.  A  cylinder  and  a  ball  each  of  radius  R  start  from  rest  and  roll 
down  an  inclined  plane  without  slipping.     If  V  is  the  velocity  of  trans- 
lation after  descending  through  a  vertical  distance  N.  show  that 

F2  =  $(2g-/i)  in  the  case  of  the  cylinder, 
and 

J72  =  %(2gh)  in  the  case  of  the  ball. 

108.  A  wheel  having  an  initial  velocity  of  10  ft.  per  sec.  ascends  an 
incline  of  i  in  100.     How  far  will  the  wheel  run  along  the  incline,  neg- 
lecting friction  ?  Ans.  232.9  ft. 

109.  A  wrought-iron  fly-wheel  10  ft.  in  diameter  makes  63  revolutions 
per  minute.     Find  the  intensity  of  stress  on  a  transverse  section  of  the 
rim,  disregarding  the  influence  of  the  arms.     If  the  wheel,  which  weighs 
W  Ibs.,  gives  out  work  equivalent  to  that  done  in  raising  W  through  a 

height  of  5i  ft.  in  i  sec.,  what  velocity  will  it  lose  ?    If  the  axle  of  the  wheel 
is  10  in.  in  diameter  and  if  .08  is  the  coefficient  of  friction,  show  that  it 

W 

will  take H.  P.  to  turn  the  wheel. 

2500 

Ans.  16,335  Ibs. ;  3.6  ft.  per  sec. 

1 10.  If  the  earth  be  assumed  to  be  spherical,  how  much  heat  would 
be  developed  if  its  axial  rotation  were  suddenly  stopped,  a  unit  of  heat 
corresponding  to  778  ft.-lbs.  ? 


EXAMPLES:  227 

Weight  of  mass  of  earth  =  io21  x  6.029  tons;  diameter  of  earth 
=  8000  miles. 

in.  A  body  weighing  50  Ibs.  is  projected  along  a  rough  horizontal 
plane,  the  velocity  of  projection  being  100  ft.  per  sec.  What  amount  of 
\vork  will  have  been  expended  when  the  body  comes  to  rest  ? 

If  the  coefficient  of  friction  is  i,  how  much  work  is  done  against  fric 
i  ion  in  4  sees.,  and  in  what  time  will  the  body  come  to  rest  ? 

Ans.  7763.9  ft. -Ibs.  ;  2298!  ft.-lbs  ;  24115- sees. 

112.  A  chain  /  ft.  in  length  and  a  sq.  in.  in  sectional  area  has  one 
end  securely  anchored,  and  suddenly  checks  a  weight  of  W  Ibs.  attached 
to  the  other  end,  and  moving  with  a  velocity  of  V  ft.  per  sec.  away  from 
the  anchorage.  Find  the  greatest  pull  upon  the  chain. 

faEW 


Ans.  Pull  =  V\ 

o 

113.  Apply  this  result  to  the  case  of  a  wagon  weighing  4  tons  and 
worked  from  a  stationary  engine  by  a  rope  3  sq.  in.  in  sectional  area. 
The  wagon  is  running  down  an  incline  at  the  rate  of  4  miles  an  hour, 
and,  after  600  ft.  of  rope  have  been  paid  out,  is  suddenly  checked  by  the 
stoppage  or  reversal  of  the  engine  (E  =  15,000,000  Ibs.). 

Ans.  26,884  Ibs. 

1 14.  A  chain  /  ft.  in  length  and  a  sq.  in.  in  sectional  area  has  one  end 
attached  to  a  weight  of  W  Ibs.  at  rest,  and  at  the  other  end  is  a  weight 
of  n  W  Ibs.  moving  with  a  velocity  of  V  ft.  per  second  and  away  from 
the  first.     Find  the  greatest  pull  on  the  chain. 

Ans.  Pull  =  Vi/  aEWn  . 
Y   lg(n  +  i) 

115.  A  dead  weight  of  io  tons  is  to  act  as  a  drag  upon  a  ship  to 
which  it  is  attached  by  a  wire  rope  150  ft.  in  length  and  having  an  effec- 
tive sectional  area  of  8  sq.  in.     If  the  velocity  of  the  floating  ship  is 
20  ft.  per  second,  and  if  its  inertia  is  equivalent  to  a  mass  of  390  tons, 
find  the  greatest  pull  on  the  chain  (E  =  15,000,000  Ibs.). 

Ans.  208  tons. 

116.  (a)  A  train  weighing  160  tons  (of  2240  Ibs.)  travels  at  30  miles 
an  hour  against  a  resistance  of  io  Ibs.  per  ton.     What  H.  P.  is  exerted  ? 

(b)  With  the  same  H.  P.  what  will  be  the  speed  up  a  gradient  of  i 
in  loo? 

(c)  If  the  steam  is  shut  off,  how  far  will  the  train  run  before  stopping 
(i)  on  the  incline;  (2)  on  the  level? 

(d)  If   the  draw-bar  suddenly  breaks,   in  what  distance  would  the 
carriages  (100  tons  in  weight)  be  stopped  if  the  brakes  are  applied  im- 
mediately the  fracture  occur?,  the  weight  of  the  brake-van  being  20  tons 
and  the  coefficient  of  friction  .2? 


228  THEORY  OF  STRUCTURES. 

(e)  If  the  engine  (weight  =  60  tons)  continued  to  exert  the  same 
power  after  the  fracture,  what  would  be  its  ultimate  speed  ? 

(/)  What  resistance  would  be  required  to  stop  the  whole  train  after 
steam  is  shut  off,  in  1000  yards  on  the  level  ? 

Ans.  (a)  128;  (b)  9^  miles  per  hour;  (c)  (i)  199.2  ft., 
(2)  6776  ft.  ;  (d)  680.3  ft.  on  the  level,  52.9  ft. 
on  the  incline ;  (e)  80  miles  an  hour  on  the 
level,  24.6  miles  on  the  incline  ;  (/)  22.58  Ibs. 
per  ton. 

117.  A  4-in.  X  3-in.  diameter  crank-pin  is  to  be  balanced   by  two 
weights  on  the  same  side  of  the  crank  ;  the  length  of  the  crank  is  12  in. ; 
the  engine  makes  100  revolutions  per  minute;  the  distance  of  the  C.  of 
G.  of  each  weight  from  the  axis  of  the  shaft  is  6  in.     Find  the  weights. 

118.  A  shaft  is  worked  with  cranks  at  120°.     Assuming  the  pressure 
on  the  crank-pin  to  be  horizontal  and  constant  in  amount,  compare  the 
coefficients  of  actual  and  ultimate  strength  to  be  used  in  calculating  the 
diameter  of  the  shaft.  Ans.  a'  =  .507^. 

119.  In  a  horizontal  marine  engine  with  two* cranks  at  right  angles 
distant  8  ft.  from  one  another,  weight  of  reciprocating  parts  attached  to 
each  crank  is  10  tons,  revolutions  75  per  minute,  stroke  4  ft.     Find  the 
alternating  force  and  couple  due  to  inertia. 

Ans.  54.2  tons;  216.8  ft.-tons. 

120.  An  inside-cylinder  locomotive  is  running  at  50  miles  an  hour; 
the  driving-wheels  are  6  ft.  in  diameter;  the  distance  between  the  centre- 
lines of  the  cylinders  is  30  in.,  the  stroke  24  in.,  the  weight  of  one  piston 
and   rod  300  Ibs.,  and  the  horizontal   distance   between   the   balance- 
weights  4!  ft. ;  the  diameter  of  the  weight-circle  is  4i  ft.     Find  the 
alternating  force  and  couple,  and  also  the  magnitude  and  position  of 
suitable  balance- weights. 

Ans.  7871  Ibs;  9839  ft.-lbs.  ;  106.5  Ibs.  5  27f°* 

121.  The  pressure  equivalent  to  the  weight  of  the  reciprocating  parts 
of  an  engine  is  3  Ibs.  per  sq.  in. ;  the  stroke  is  36  in.  ;  the  number  of 
revolutions  per  minute  is  45  ;  the  back-pressure  is  2  Ibs.  per  sq.  in.;  the 
absolute  initial  steam- pressure  is  60  Ibs.  per  sq.  in.;  the  rate  of  expansion 
is  3.     Find  the  pressure  necessary  to  start  the  piston,  and  also  the  effec- 
tive pressure  at  each  -J-  of  the  stroke. 

122.  An  engine  with  a  24-in.  cylinder  and  a  connecting-rod  =  six  cranks 
=  6  ft.,  makes  60  revolutions  per  minute.     Show  that  the  pressure  re- 
quired to  start  and  stop  the  engine  at  the  dead-points  =  ^  of  the  weight 
of  reciprocating  parts. 

123.  Find  the  ratio  of  thrust  at  cross-head  to  tangential  effort  on 
crank-pin  when  the  crank  is  45°  from  the  line  of  stroke,  the  connecting- 
rod  being  =  four  cranks. 


EXAMPLES,  229 

124.  Draw  the  linear  diagram  of  crank-effort  in  the  case  of  single 
crank,  the  connecting-rod  being  =  four  cranks.     Assume  the  resistance 
uniform  and  a  constant  pressure  of  9000  Ibs.  on  the  piston,  the  stroke 
oeing  4  ft.  and  the  number  of  revolutions  per  minute  55.     Also  find  the 
ductuation  of  energy  in  ft.-lbs.  for  one  revolution. 

125.  An  engine  with  a  connecting-rod  =  six  cranks  =  6  ft.  receives 
steam  at  70  Ibs.  pressure  per  sq.  in.,  and  cuts  off  at  one-quarter  stroke. 
Find  the  crank-effort  when  the  piston  has  travelled  one  third  of  its  for- 
ward stroke.     Diameter  of  piston  =  2  ft.     Also  find  the  position  of  the 
piston  where  its  velocity  is  a  maximum. 

126.  Data:  Stroke  =  3  ft.  ;  number  of  revolutions  per  minute  =  60; 
cut-off  at  one-half  stroke;  jnitial  pressure  =  56  Ibs.  per  sq.  in.  absolute; 
diameter  of  piston  =  10  in. ;  weight  of  reciprocating  parts  =  550  Ibs.  ; 
back-pressure  =  i\  Ibs.  per  sq.  in.  absolute.     Find  the  effective  pressure 
at  each  fourth  of  the  stroke,  taking  account  of  the  inertia  of  the  piston. 
Also   find   the   pressure   equivalent    to   inertia   at  commencement  of 
stroke. 

127.  A  pair  of  250  H.  P.  engines,  with  cranks  at  90°,  and  working 
against  a  uniform  resistance  and  under  a  uniform  steam-pressure,  are 
running  at  60  revolutions  per  minute.     Assuming  an  indefinitely  long 
connecting-rod,   find  the  maximum  and  minimum  moments  of  crank- 
effort,  the  fluctuation  of  energy,  and  the  coefficient  of  energy. 

128.  An  inside-cylinder  locomotive  runs  at  25  miles  per  hour;  its 
drivers  are  60  in.  in  diameter;  the  stroke  is  24  in. ;  the  distance  between 
tne  centre-lines    of   the   cylinders  =  30  in. ;    weight   of   reciprocating 
parts  =  500  Ibs. ;  horizontal  distance  between  balance-weights  =  59  in.  ; 
diameter  of  weight-circle  =  42  in.     Find  the  alternating  force,  alternat- 
ing couple,  and  the  magnitude  and  position  of  suitable  balance-weights. 

Ans.  226.8  Ibs.;  4113.8  ft.-lbs. ;  0  =  26°. 

129.  Draw  a  diagram  of  crank-effort  for  a  single  crank,  the  connect- 
ing-rod being  equal  to  four  cranks,  the  stroke  4  ft.,  and  the  number  of 
Devolutions  per  minute  55.     Assume  a  uniform  resistance  and  a  constant 

•ressure  of  9000  Ibs.  on  the  piston. 

130.  A  vertical  prismatic  bar  of  weight  £Fi ,  sectional  area  A,  and 
length  L  has  its  upper  end  fixed,  and  carries  a  weight  W*  at  the  lower 
end.     Find  the  amount  and  work  of  the  elongation. 

Ans.  Ext.  =  J^(-Y  +  W*\  ;  work  =  *g( —  +  ^  ^  + 

131.  A  right  cone  of  weight  W  and  height  h  rests  upon  its  base  of 
radius  r.     Find  the  amount  and  work  of  the  compression. 

.        _  Wh  i   W* 

Ans.  Comp.  =  — —  ;  work  =  ~  — =-.. 
2itEr*  8  Tthr* 

132.  A  tower  of  height  h,  in  the  form  of  a  solid  of  revolution  about  a 
vertical  axis,  carries  a  given  stir  charge.     If  the  specific  weight  cf  the 


230  THEORY  OF  STRUCTURES. 

material  of  the  tower  is  w,  and  the  radius  of  the  base  a,  determine  the 
curve  of  the  generating  line  so  that  the  stress  at  every  point  of  the  tower 
may  be/.  If  the  surcharge  is  zero  and  the  height  of  the  tower  becomes 
infinite,  show  that  its  volume  remains  finite. 

_-wx  f 

Ans.  y  =  ae     *f  ;  vol.  of  tower  of  infinite  height  =  —  fta*. 

w 

133.  Determine   the  generating  curve  when  the  tower  in  the  last 
question  is  hollow,  the  hollow  part  being  in  the  form  of  a  right  cylinder 
upon  a  circular  base  of  given  radius  R. 

Ans.  y*  -  K*  =  (a?  -  K')e  ~  T. 

134.  A  heavy  vertical  bar  of  length  /  and  specific  weight  w  is  fixed 
at  its  upper  end  and  carries  a  given  weight  Wat  the  lower  end.     Deter- 
mine the  form  of  the  bar  so  that  the  horizontal  sections  may  be  pro- 
portionate to  the  stress/  to  Which  they  are  subjected.     (Note.  —  Such  a 
bar  is  a  bar  of  uniform  strength.) 

W,  J(l~x] 
Ans.  Sectional  area  at  distance  x  from  origin  =  —  e 

135.  Find  the  upper  and  lower  sectional  areas  of  a  steel  shaft  of  uni- 
form strength,  200  ft.  in  length,  which  will  safely  sustain  its  own  weight 
and  100  tons,  7  tons  per  sq.  in.  being  the  working  stress. 

Ans.   14.3  sq.  in.  ;  17.8  sq.  in. 

136.  A  vertical  elastic  rod  of  natural  length  L  and  of  which  the  mass 
may  be  neglected,  is  fixed  at  its  upper  end  and  carries  a  weight  W\  at 
the  lower  end.     A  weight  W2  falls  from  a  height  k  upon  W\  .     Find  the 
velocity  and  extension  of  the  rod  at  any  time  /, 


¥  ")-(£)' 


x  being  measured  from  mean  position  of  (  W\  +  Wa). 

137.  Determine  the  functions  /''and/  in  Art.  24  when  P\  is  zero,  and 
also  when  the  rod  is  perfectly  free  ;  i.e.,  when  Po  =  o  and  Pi  =  o. 

138.  An  elastic  trapezoidal  lamina  ABCD,  of  natural  length  /  and 
thickness  unity,  has   its  upper  edge  AB  (20)  fixed  and   hangs  verti- 
cally.    If  a  weight  Wis  suspended  from  the  lower  edge  CD  (2$),  show 
that,  neglecting  the  weight  of  the  lamina,  the  consequent  elongation 

=  --TT  —  ^7~ll°g«T-     If  an  additional   weight  is  placed  upon   W  and 
then  suddenly  removed,  show  that  the  oscillation  set  up  is  isochronous 


and  that  the  time  of  a  complete  oscillation  =  n\  — -=r 7-  f . 

Examine  the  case  when  a  =  b. 

Ans.  Ext.  = ;  time  of  oscillation  =  7r,4/____ 

2<*£  Y 


EXAMPLES.  231 

139.  If  the  specific  weight  of  the  lamina  in  the  preceding  question  is 
w,  find  how  much  it  will  stretch  under  its  own  weight,  and  also  the 
work  of  extension.     Determine  the  result  when  a  —  b. 

i      wbT  b       wra  +  d     wl* 

AnS'   2~E  ^7  10g  ~a  +  iff  ^  ;  2^' 
ivlz         (       «4  —  b*       2,0  a  }     "dFal* 

=  ^r^Jr  \  -  —  +  f(a  ~  d>-  b  !og<  ?  \  ''  -W- 

140.  An  elastic  lamina  in  the  form  of  an  isosceles  triangle  ABC  has 
its  base  AB  (=  20)  fixed  and  hangs  vertically.     If  its  weight  is  W,  find 
its  elongation.     Take  coefficient  of  elasticity  =  £,  thickness  of  lamina 

=  unity,  and  L  the  distance  of  C  from  AB.  WL 

Ans.  -- 


141.  A  metal  rod  £•  sq.  in.  in  area  and  5  ft.  long  hangs  vertically  with 
its  upper  end  fixed  and  carries  a  weight  of  18  Ibs.  at  the  lower  end.     On 
striking  the  rod  it  emitted  a  musical  note  of  264  vibrations  per  second 
(middle  C  of  piano-forte).     Find  the  coefficient  of  elasticity,  the  weight 
of  the  rod  being  neglected.  Am.  30,979,160  Ibs. 

142.  Diameter  of  a  pipe  is  18  in.  ;  at  one  point  it  is  curved  to  an  arc 
of  6  ft.  radius.     Water  flows  round  the  curve  with  a  velocity  of  6  ft.  per 
second.     Determine  the  centrifugal  force  per  foot  of  length  of  elbow- 
measured  along  the  axis.  Ans.  124.3  Ibs. 

143.  A  disk  of  weight  £Fand  area  A  sq.  ft.  makes  n  revolutions  per 
second  about  an  axis  through  its  centre,  inclined  at  an  angle  0  to  the 
normal  to  the  plane  of  the  disk.     Find  the  centrifugal  couple. 

WAn* 

Ans.  --  tan  6  ft.  -Ibs. 
5.12 

144.  In  a  circular  pipe  of  internal  radius  r  and  thickness  /,  a  column 
of  water  of  length  /,  flowing  with  a  velocity  due  to  the  head  h,  is  sud- 
denly checked.     Show  that 


E  being  the  coefficient  of  elasticity  of  the  material  of  the  pipe,  E\  the 
coefficient  of  compressibility  of  the  water,  and  A  the  extension  of  the 
pipe  circumference  corresponding  to  E. 

145.  A  heavy  ball  attached  by  a  string  to  a  fixed  point  O  revolves  in 
a  horizontal  circle  with  a  given  uniform  angular  velocity  GO.  Find  the 
vertical  depth  of  the  centre  of  the  ball  below  the  point  of  attachment. 

If  a  uniform  rod  be  substituted  for  the  ball  and  string,  find  its 
position. 

Also  find  the  position  when  the  ball  is  attached  to  the  fixed  point  by 


232  THEORY   OF  STRUCTURES. 

a  uniform  rod  ;  r  being  the  ratio  of  the  weight  of  the  rod  to  the  weight 

of  the  ball. 

n 

i  +  — 

GO"*  '  i  c«32 '  Gfl2          n 

i  +  — 

3 

146.  The  deflection  of  a  truss  of  /  ft.  span  is  /  x  .001  under  a  station- 
ary load  W.    What  will  be  the  increased  pressure  due  to  centrifugal  force 
when  W7 crosses  the  bridge  at  the  rate  of  60  miles  an  hour?        ..-,  , 

242  W 
Ans.  -  -  — r-. 
125    / 

147.  A  fly-wheel  20  ft.   in  diameter  revolves  at  30  revolutions  per 
minute.     Assuming  weight  of  iron  450  Ibs.  per  cu.  ft.,  find  the  intensity 
of  the  stress  on  the  transverse  section  of  the  rim,  assuming  it  unaffected 
by  the  arms.  Ans.  96  Ibs.  per  sq.  in. 

148.  Assuming  15,000  Ibs.  per  sq.  in.  as  the  tensile  strength  of  cast- 
iron,  and  taking  5  as  a  factor  of  safety,  find  the  maximum  working  speed 
and  the  bursting  speed  for  a  cast-iron  fly-wheel  of  20  ft.  mean  diameter 
and  weighing  24,000  Ibs.,  the  section  of  the  rim  being  160  sq.  in. 

149.  A  6o-in.  driving-wheel  weighs  3^  tons,  and  its  C.  of  G.  is  i  in. 
out  of  centre.     Find  the  greatest  and  the  least  pressure  on  the  rails. 

150.  A   wheel   of  weight  W,   radius   of  gyration  k,   and   making  n 
revolutions  per  second  on  an  axle  of  radius/?,  comes  to  rest  after  having 
made  N  revolutions.     Find  the  coefficient  of  friction. 

Ans.  sin  <f>  = ,  and  coeff.  of  fric.  =  tan  0. 

Ng 

151.  A  train  starts  from  a  station  at^f  and  runs  on  a  level  to  a  station 
at  B,  I  ft.  away.     If  the  speed  is  not  to  exceed  v  ft.  per  sec.,  show  that 
the  time  between  the  two  stations  is 

l_       Wv          P  +  B 

"r 


•v       g  2  (P-R)(B  +  R)' 

W  being  the  gross  weight  of  the  train,  P  the  mean  uniform  pull  exerted 
by  the  engine,  R  the  road  resistance,  and  B  the  retarding  effect  of  the 
brakes. 

Also,  if  the  speed  is  not  limited,  show  that  the  least  time  in  which 
the  train  can  run  between  the  specified  points  is 


^    W_         P  +  B 
'  ~i  (P  —  R)(B+R) 
and  that  the  maximum  speed  attained  is 


w     ?,s--*-v«~* 


EXAMPLES.  233 

152.  A  locomotive  capable  of  exerting  a  uniform  pull  of  2  tons,  with 
a  24-in.  stroke,  2o-in.  cylinder,  and  6o-in.  driving-wheels,  hauls  a  train 
between  two  stations  3  miles  apart.     The  gross  weight  of  the  train  and 
locomotive  =  200  tons;  the  road  resistance  =  12  Ibs.  per  ton  (of  2000 
Ibs.) ;  the  brakes,  when  applied,  press  with  two  thirds  of  the  weight  on 
the  wheels  of  the  engine  and  brake-van,  viz.,  90  tons,  the  coefficient  of 
friction  being  .18.     Find  (a)  the  least  time  between  the  stations;  (b)  the 
distance  in  which  the  train  is  brought  to  rest;  (c)  the  maximum   speed 
attained  ;  (d)  the  pressure  of  the  steam  ;  (e)  the  weight  upon  the  driving- 
wheels. 

Ans. — 0)  513.8  sec.  ;  (6)  990  ft.;  (c)  42  miles  per  hour;  (d)   25 
Ibs.  per  sq.  in. ;  (<?)  i  \\  tons. 

153.  If  the  speed  in  the  last  question  is  limited  to  30  miles  an  hour, 
find  (a)  the  time  between  the  stations;  (£)the  distance  in  which  the  train 
is  brought  to  rest ;  (c)  the  distance  traversed  at  30  miles  an  hour. 

Ans.— (a)  543i  sec. ;  (b)  504^  ft. ;  (c)  7773^  ft. 

154.  If  the  steam-pressure  in  the  above  locomotive  is  increased  to  50 
Ibs.  per  sq.  in.,  find  (a)  the  weight  of  the  heaviest  train  which  can  be 
hauled  between  the  stations  in   10  minutes,  the  road-resistance  being  20 
Ibs.  per  ton  (of  2000  Ibs.)  and  the  braking  power  being  sufficient  to  bring 
the  train  to  rest  in  a  distance  of  720  ft. 

Also  find  (b)  the  braking  power;  (c)  the  weight  thrown  upon  the 
drivers,  the  coefficient  of  friction  being  % ;  (d)  the  maximum  speed 
attained. 

Ans. — (a)  310^ tuns;  (b)  15. 6  tons;  (c)  24  tons;  (d)  36  miles  per  hour. 

155.  The  weight  upon  the  driving-wheels  (D  in.  in  diameter)  of  a 
locomotive  is  W  tons;  the  adhesion  =  one  fifth  ;  the  cylinders  have  a 
diameter  of  d  in.   and  a  stroke  of  /  in.     Find  the  steam-pressure  re- 
quired to  skid  the  wheels.  IV D 

Ans.  400-——  Ibs.  per  sq.  in. 

156.  Two  trains,  each  with  a  brake-power  of  190  Ibs.  per  ton  (of  2000 
Ibs.),  run  between  Montreal  and  Toronto,  a  distance  of  333  miles,  against 
an  average  resistance  of  10  Ibs.  per  ton.     One  train  runs  through,  and 
the  other  stops  at  N  intermediate  stations.     Show  that  the  saving  of 

QN 

fuel  in  the  former  is  —  per  cent ;  the  speed  is  not  to  exceed  30  miles 

per  hour. 

157.  If  the  end  of  a  railway  wagon  exposes  a  surface  of  6  x  4  ft.  to 
the  wind,  what  is  the  greatest  gradient  up  which  a  20  Ib.  to  the  sq.  ft. 
gale  will  drive  it?     Take  the  weight  at  10  tons,  the  friction  10  Ibs.  per 
ton.  Ans.    I  in  59. 

1 58.  A  locomotive  and  tender  weigh  70  tons,  of  which  26  tons  are  car- 
ried by  the  driving-wheels.     Taking  the  adhesion  at  £,  friction  10  Ibs.  per 
ton,  what  maximum  gradient  can  the  engine  ascend?         Ans.   I  in  16. 


234  THEORY  OF  STRUCTURES. 

1  59.  Given  a  locomotive  with  two  18"  x  26"  cylinders,  the  connecting- 
rod  =  6  ft.,  the  boiler-pressure  =  140  Ibs.,  and  driving-wheels  of  7'  o" 

force  at  periphery 

diameter,  calculate  the  adhesion-friction,  i.e.,  the  ratio  —  —  --  —  :  -  . 

weight  on  drivers 

160.  A  railway  wagon  weighing  20  tons,  with  two  pairs  of  wheels 
8'  o"  centre  to  centre,  and  with  its  centre  of  inertia  7'  o"  above  top  of 
rails,  has  its  wheels  skidded  while  running.     Take  jn  =  0.15.     Required 
the  total  retarding  force  and  pressure  of  each  wheel. 

Ans.  7.375  ;  12.625,  and  3  tons  on  rail. 

161,  Find  (a)  the  least  time  in  which  a  locomotive  exerting  a  uniform 
pull  of  P  tons  can  haul  a  train  weighing  W  tons  between  two  stations 
/  ft.  apart  on  an  incline  of  I  in  m,  the  brake-power  being  B  tons  and  the 
road-resistance  R  tons. 

Also  find  (V)  the  time  between  stations  when  the  speed  is  limited  to 
v  ft.  per  sec. 

p  +  £  l       Wv          P  +  B 


W 

where  A  —  R  +  —  . 
m 

162.  A  locomotive  exerting  a  uniform  pull  of  4  tons  hauls  a  train  of 
200  tons  up  an  incline  of  i  in  200,  between  two  stations  2  miles  apart, 
the  greatest  allowable  speed  being  30  miles  an  hour.     If  the  road-resist- 
ance is  10  Ibs.  per  ton  (of  2000  Ibs.),  and  if  the  brakes  are  capable  of  ex- 
erting a  pressure  of  100  tons,  the  adhesion  being  one  fifth,  find  (a)  the 
time  between  the  stations;  (b)  the  distance  in  which  the  train  is  brought 
to  rest;  (c)  the  distance  traversed  at  30  miles. 

Also,  if  the  speed  is  not  limited  to  30  miles,  find  (d)  the  least  time  in 
which  the  distance  can  be  accomplished  ;  (e)  the  maximum  speed  attained  ; 
(/)  the  distance  in  which  the  train  is  brought  to  rest. 

Ans.—  (a)  5i  min.  ;    (ff)  275  ft.;    (c)  7260  ft.;   (d)  4.87  min.; 
(e)  53.8  miles  per  hour;  (/)  880  ft. 

163.  With  the  same  brake-power,  adhesion,  and  road-resistance,  find 
the  weight  of  the  heaviest  train  which  the  locomotive  in  the  preceding 
question,  exerting  the  uniform  pull  of  4  tons,  can  haul  between  the  two 
stations  in  6  minutes.  Ans.  360  tons. 

164.  If  the  locomotive  has  6o-in.  drivers  and  24-in.  x  2o-in.  diameter 
cylinders,  find  the  weight  required  upon  the  drivers  when  the  steam- 
pressure  is  50  Ibs.  per  sq.  in.  Ans.  20  tons. 


CHAPTER    IV. 

STRESSES,  STRAINS,  EARTHWORK  AND  RETAINING- 

WALLS. 

1.  Internal  Stresses. — The  application  of  external  forces 
to  a  material  body  will  strain   or  deform  it,  and  the  particles 
of  the  body  will  be  in  a  state  of  mutual  stress. 

In  the  following  calculations  it  is  assumed  : 

(a)  That  the  stresses  under  consideration  are  parallel  to  one 
and  the  same  plane,  viz.,  the  plane  of  the  paper. 

(b)  That  the  stresses  normal  to  this  plane  are  constant  in 
direction  and  magnitude. 

(c)  That  the  thickness  of  the  plane  is  unity. 

Def.  The  angle  between  the  direction  of  a  given  stress  and 
the  normal  to  the  plane  on  which  it  acts  is  called  the  obliquity 
of  the  stress. 

2.  Simple  Strain. — The  solid  ABCD  (Fig.  207)  of  uniform 
transverse  section  A  is  acted  upon  in  the  direction  of  its  length 

by  a  force  P  uniformly  distributed  over  its  end, 

p 
producing  an  intensity  of  stress  — -  =/.     At  any  |P 


m 


k/ 

\o/ 

-j/ 


other  transverse  section  mn  the  intensity  must  be 
the  same  in  order  that  equilibrium  may  be  main- 
tained. 

Draw  an   oblique  plane   m'n' ,  inclined  at  an 
angle  6  to  the  axis.     The  total  stress  on  m'n'  =  P  ? 
and  necessarily  acts  in  the  direction  of  the  axis.   m 

p 

The   intensity  of    the   stress  on   m'n'  =  = 

m'n' 
p  p 

a  =  -T  sin  6  =  p  sin  6.    The  normal  com-        FlG-  2°7- 

mn  cosec  0      A 

ponent  of  the  intensity  on  m'n'  =  p  sina  6  =  pn'. 

235 


c 


236  THEORY   OF  STRUCTURES. 

The  tangential  component  or  shear  on  m'ri 
—  p  sin  8  cos  8  =  //. 

^•0,  if  m"n"  is  an  oblique  plane  perpendicular  to  m'n' ,  the  nor- 
iiial  component  of  the  intensity  on  m"n"  =  p  cos2  8  =  pn". 
The  tangential  component  or  shear  on  m"n" 

=  p  cos  8  sin  8  =  p" . 
...  A'  +  p"  =  p      and  p{  =  p;'  =  p  sin  8  cos  0  =  ^5ll£?. 


The    shear  is   evidently    a   maximum    when    28  =  90°    or 

9  =  45°. 

3.  Compound  Strain. — (a)   First  consider  an  indefinitely 
small  rectangular  element  OACB  (Fig.  208)  of  a  strained  body, 
p        „       p      p  kept    in    equilibrium    by   stresses 

\       \v      \      \  acting  as  in  the  figure. 

^ ^ ^ L — £  /  is  the  intensity  of  stress  on 

\^  the  faces  OB,  AC,  and  a  its  ob- 
x^g  liquity. 

0     \       \       \x      \    B  q  is  the  intensity  of  stress  on 

\      \       \      \          the  faces  OA,  BC,  and  /?  its  ob- 

FIG.  208.  liquity. 

OB  .p  cos  a,  the  total  normal  stress  on  OB,  is  balanced  by 
AC  .p  cos  a,  the  total  normal  stress  on  AC. 

OB  .p  sin  a,  the  total  shear  on  OB,  is  equal   in   magnitude 
but  opposite  in  direction  to  AC  .p  sin  a,  the  total  shear  on  AC. 
These    two    forces,  therefore,    form    a   couple   of   moment; 
JB  ./sin  ex .  OA. 

Similarly,  the  total  normal  stresses  on  the  faces  OA, 
BC  balance  and  the  total  shears  form  a  couple  of  moment 
OA  .  q  sin  /3 .  OB. 

In  order  that  equilibrium  may  be  maintained  the  two 
couples  must  balance. 

.-.  OB  .p  sin  a  .  OA  =  OA  .  q  sin  /? .  OB, 

or 

/  sin  a  —  q  sin  ft  —  /,  suppose. 


COMPOUND    STRAIN. 


237 


Hence,  at  any  point  of  a  strained  body,  the  intensities  of  the 
shears  on  any  two  planes  at  right  angles  to  each  other  are  equal. 

(b]  Next  consider  an  indefinitely  small  triangular  element 
OAB  (Fig.  209)  of  the  strained  body,  bounded  'c 
by  a  plane   AB  and  two  planes  OA,  OB  at 
right  angles  to  each  other. 

Let  /  be  the  intensity  of  stress  on  OB, 
a  its  obliquity. 

Let   q  be  the  intensity  of  stress  on  OA, 
/3  its  obliquity.  /\ 


Let  /  be  the  intensity  of  shear  on  each  of     0     B  D 

the  planes  OA,  OB.     Then  FlG- 2°9' 

/  —  p  sin  a  =  q  sin  ft. 

pn ,  the  normal  component  of  p,  =  p  cos  a. 
qn ,     •'          "  "  "  q,  =  q  cos  ft. 

Produce  OA  and  take  OC  =  pn  .  OB  +  t .  OA  =  the  total 
force  on  OB  in  the  direction  of  OA. 

Produce  OB,  and  take  OD  —  qn  .  OA  +  t .  OB  —  the  total 
force  on  OA  in  the  direction  of  OB. 

Complete  the  rectangle  CD. 

OE  represents  in  direction  and  magnitude  the  resultant  of 
the  two  forces  OC,  OD,  and  must  therefore  be  equal  in  magni- 
tude and  opposite  in  direction  to  the  total  stress  on  AB. 

Let/r  be  the  intensity  of  stress  on  AB.     Then 

(pr .  AB}*  =  OE*  =  OC*  + 


.fOB\*         JOAy 


.OA? 
+  (qn  .OA+t.  03?  ; 

OA  .  OB 

or     A-  =  A'  <v-^j  +  q:  \j-B)  +  **  -jff-i  Pn  +  ft)  T  • ; 

Let  x  be  the  angle  between  y4.#  and  OA.     Then, 
/V2  =  /wa  sin2  y  +  ?M2  cosa  y  +  2t  sin  r  cos  7  (pn  +  qn)  +  /a. 

This  gives  the  intensity  of  stress  on  any  plane  AB  inclined 
at  an  angle  y  to  OA,  and  in  the  limit  AB  is  a  plane  through  O. 


238  THEORY  OF  STRUCTURES. 

EXAMPLE.  —  Consider  an  indefinitely  small   triangular  ele- 

ment abc  (Fig.  210)  of  a  horizontal  beam  bounded  by  a  plane 

-  a  —    —  b   t.ab*8.db  —    be  inclined  at  0  to  the  vertical,  the 

/'^  horizontal  plane  ab,  and  the  ver- 

' 


ac. 
tf  The  element  abc  is  kept  in  equ 


FIG.  210.  librium  by  the  stress/  .  ac  upon  ac, 

the  shear  s.ab  (=  t.ab)  along  ad,  the  shear  t  .ac  along.  ac,  and 
the  stress  developed  in  the  plane  be.  The  weight  of  the  element 
is  neglected  as  being  indefinitely  small  as  compared  with  the 
forces  to  which  it  is  subjected.  Let  the  stress  upon  be  be 
decomposed  into  two  components,  the  one  X  '.  be  normal  and 
the  other  Y.  be  tangential  to  be. 

Resolving  perpendicular  and  parallel  to  fc, 

X  .be  =  p  .ac  cos  0  —  t  .  ab  cos  0  —  t  .  ac  sin  6 
and 

Y.  be  =  p  .  ac  sin  6  —  t  .  ab  sin  B  +  t  .  ac  cos  6, 
or 

X  =  PCQS*  6  —  t  sin  20  .     .......     (i) 

and 


The  value  of  B  for  which  X  is  a  maximum  is  given  by 
-      =  o  =  —  /  sin  26  —  2t  cos  20,     or     tan  26  =  --  .     (3) 
Substituting  the  value  of  B  in  eq.  I,  we  have 

max.  value  of  X  =  £-  +  A  /—  +  f.      .      .     (4) 

The  value  of  B  for  which  Y  is  a  maximum  is  given  by 

dY  P 

—  —  o  =/  cos  2/9  —  2t  sin  20,     or     tan  2B  =       .       (5) 


COMPOUND    STRAIN.  239 

Substituting  the  value  of  6  in  eq.  (2),  we  have 


max.  value  of  Y  =  \/-  +  t* ,     (6) 


Eq.  (4)  gives  the  maximum  intensity  of  stress  of  the  same 
kind  as  p.     The  maximum  intensity  of  the  opposite  kind  of 


stress 

o.         x/    4 

Eq.  (6)  gives  the  maximum  intensity  of  shear. 

The  position  of  the  planes  of  principal  stress  (see  following 

2t 

article)  is  given  by  tan  20  —  — . 

Let  0j ,  02  be  the  values  of  0  for  which  X  and  Kare  respec- 
tively maxima.     Then 


tan  20,  tan  20,=  -'J*?** 
and 

•••  *,-*,  =  45°- 

Hence,  at  any  point,  the  angle  between  the  plane  upon 
which  the  normal  intensity  of  stress  is  a  maximum  and  the 
plane  upon  which  the  tangential  intensity  of  stress  is  a  maxi- 
mum, is  equal  to  45°.  • 

Again,  t  is  zero  when  0,  =  90°  or  o°,  and  p  is  zero  when 

*,  =  45°. 

Thus,  the  curve  of  greatest  normal  intensity  cuts  the  neutral 
axis  at  an  angle  of  45°,  one  skin  surface  at  90°  and  the  opposite 
at  o°,  while  the  curve  of  greatest  tangential  intensity  cuts 
the  skin  surfaces  at  45°,  and  touches  the  neutral  axis. 

Fig.  211  serves  to  illustrate  the  curves  of  greatest  normal 
intensity.  There  are  evidently  two  sets  of  these  curves,  re- 
ferring respectively  to  direct  thrust  and  direct  tension. 


240  THEORY   OF  STRUCTURES. 

Fig.  212    illustrates   the   curves  of   greatest  tangential  in- 
tensity. 


V- 


FIG.  211.  FIG.  212. 

4.  Principal  Stresses. — Suppose   that   there  is  no  shear 
on  AB,  Fig.  209,  and  that  the  stress  is  wholly  normal. 
In  such  a  case  OE  must  be  perpendicular  to  AB. 


OC      OC      Pn-OB  +  t.OA 


qn  +  t  tan  y 


It 

qn—pn      i  —  tan*  x 


^E         Two  values  of  y  satisfy  this  equation,  viz., 
/\     y  and  y  +  90°. 

Hence,  at  any  point  of  a  strained  body,  there 
are  two  planes  at  right  angles  to  each  other,  on 
which  the  stress  is  wholly  normal. 

Such  planes  are  called  planes  of  principal 
stress,  and  the  stresses  themselves  principal 
stresses. 

5.  Ellipse  of  Stress. — At  any  point  of  the 
strained  body,  consider  a  small  triangular  ele- 
ment OAB  (Fig.  213),  OA  and  OB  being  the  planes  of  principal 
stress. 

Let  /,  be  the  principal  stress  normal  to  OB. 
"         "    "          "  "         "        "  OA. 


CONSTANT  COMPONENTS  OF pr. 


241 


Complete  the  construction  as  before,  and  let  $  be  the  angle 
between  OE  and  OC.     Then 


CE      OD      p,OA 
- 


A... 


•    •    •     (8) 


pr  sin  w       .,  •           pr  cos  m 
= i ,     sin  v  = ;  and 


cos       = 


A 


••    •     (9) 


Take  6^7?  to  represent  /r  in  direction  and  magnitude. 

Let  X,  Y  be  the  co-ordinates  of  R  with  respect  to  O.    Then 


=  A  cos     ,         =  /r  sn 


and  eq.  (9)  becomes 


the  equation  to  an  ellipse  with  its  centre  at  O,  and  its  axesr 
(equal  to  2/>,  and  2/>2)  lying  in  the  planes  of  principal  stress. 
This  ellipse  is  called  the  ellipse  of  stress,  and  the  stress  on  any 
plane  AB  at  O  is  the  semi-diameter  of  the  ellipse  drawn  in  a  di- 
rection making  an  angle  ip  with  the  axis  OC,  fy  being  given  by 

tan  i/j  =  —  cot  y.     (Eq.  (8).)    .     .     .     .     (n) 

6.  Constant  Components   of  pr.  —  Take   the  planes   of 
principal  stress  as  planes  of  reference  (Fig.  214). 

F 
A 


FIG. 


242  THEORY  OF  STRUCTURES. 

Draw  ON  perpendicular  to  AB,  and  take  ON  —-^ 

Let  the  obliquity  of  OR  =  0  =  RON  =  90°  -  </>  -  y. 
Join  NR.    Then 

NR  =  OR"  +  ON2  -  2OR .  ON  cos  0 

=  Pr  +  (^~}  -  A(A  +  A)  sin  (0  +  7 

But  A8  =  A2  sin2  r  +A2  cos'  x>    and 


sin  (^  +  y)  =  sin  ^'cos  y  +  cos  ^  sin  y  =  ^  cosa  y  +  --  sin"  y. 


(See  eqs.  (8).) 


.-.  NR*  =  Aa  sin8  y  +  A'  cos2  r  +  (—2 

-(A+A)(Asin8y+Acos8 


(A+AV      .  .       fA-AV 
=  V— T—  )  -AA- I     2     |- 


(12) 


Hence,  /^^  intensity  of  stress  OR  at  any  point  O  of  the  plane 
A  OB  is  the  resultant  of  two  constant  intensities 


and 


the  former  being  perpendicular  to  the  plane. 


THE  ANGLE  ONR.  243 

7.  The  Angle  ONK  =  2y. 

sin  ONR       OR  pr 

sin  0      ~  NR  ~~  pl  —  A' 

2 
But 

sin  0  =  cos  (&  +  y)  =  cos  ^  cos  y  —  sin  ^  sin  y 

=  ^~     sin  r  cos   =          sih  2* 


sn 


2A  2 

.-.  sin  OA7^?  =  sin  2y,     or     (7^7?  —  2y.     .     .     (13) 

Let  NR  (Fig.  214)  produced  in  both  directions  meet  OA  in 
F  and  O£  in  (7. 


The  angle     OFN  =  180°  -  ONR  -  NOF 

=  180°  -  2y  -  (90°  -  Y)  =  90°  -  y  =  FON. 

.'.  NF  =  NO  ;     so,     NG  =  NO  = 

.'.  A7"  is  the  middle  point  of  F£. 
Also 

RF=FN-  NR  =  ON-  NR  = 
and 

RG  =  RN+NG  =  RN+  ON  = 

N.  B.—  The  shear  at  O 
A -A 


cos  (2x  -  90°)  =  (A  -A)  sin  Y  cos  X- 


244  THEORY  OF  STRUCTURES. 

8.  Maximum  Shear. — ON  has  no  component  along  AB. 
Hence,  the  shear  on  AB  is  NR  cos  (angle  between  NR  and 
AB),  and  is  evidently  a  maximum  when  the  angle  is  nil.     Its 

value  is  then  NR,  or  Pl  ~  P\ 

9.  Application  to  Shafting. — At  any  point  in  a  plane  sec- 
tion of  a  strained  solid,  let  r  be  the  intensity  of  stress,  and  6 
its  obliquity. 

At  the  same  point  in  a  second  plane  let  s  be  the  intensity 
of  stress,  and  0'  its  obliquity. 

By  Art.  6,  r  and  s  are  the  resultants  of  two  constant 
stresses 


and 

+  /-«/,  +  A)  cos*'. 


Subtracting  one  equation  from  the  other, 


First.  Consider  the  case  of  combined  torsion  and  bending, 
as  when  a  length  of  shafting  bears  a  heavy  pulley  at  some  point 
between  the  bearings. 

Let  p  be  the  intensity  of  stress  (compression  or  tension) 
due  to  the  bending  moment  Mb. 


APPLICATION   TO   SHAFTING. 


245 


Let  q  be  the  intensity  of  shear  due  to  the  twisting  mo- 
ment Mt  . 

p  and  q  act  in  planes  at  right  angles  to  each  other. 

.-.  r  cos  6  =  py     r  sin  6  =  q  —  s,     and     6'  =  90°. 
/.  ra  =/a  +  £*     and     s  =  q. 
Hence,  by  eq.  (16), 


and  by  eq.  (15), 


and 


The  max.  shear  = 


__  A  - 


-V? 


+2*;. 


also 


(17) 


(19) 


(20) 


(21) 


=  (Chap.  VI.)    and    f  =       r      (Chap.  IX.) 


for  a  shaft  of  radius  r. 


t'\;     .    .    .    (22) 


246 
and 


THEORY  OF  STRUCTURES. 


(23) 


&==(- 


Perhaps  the  most  important  example  of  the  application  of 
the    above    principle  is  the  case  of  a 
shaft  acted  upon  by  a  crank  (Fig.  215). 
A  force  P  applied  to  the  centre  C  of 
the   crank-pin   is  resisted   by  an   equal 
and  opposite  force  at  the  bearing   JB, 
forming  a  couple  of  moment  P.  CB  =  M. 
This  couple  may  be  resolved  into  a 
FIG.  215.  bending  couple  of  moment  Mb  =  P.  AB 

=  P.  BC  cos  d  =  J/cos  tf,  and  a  twisting  couple  of  moment 
Mt  =  P  .  AC  =  P  .  BC  sin  d  =  M  sin  d ;  tf  being  the  angle 
ABC. 


',  .    .    (24) 


2M 

and  the  max.  shear  =  — ,  • 
nr 


(25) 


If  the  working  tensile  or  compressive  stress  (/>,)  and  the 
working  shear  stress  (  — ?)  are  given,  the  corresponding 

values  of  r  may  be  obtained  from  eqs.  (22)  and  (23)  or  eqs. 
(24)  and  (25) ;  the  greater  value  being  adopted  for  the  radius 
of  the  shaft. 

Second.  Consider  the  case  of  combined  torsion  and  tension 
or  compression. 

Let  the  tensile  or  compressive  force  be  P. 

P 

p,  the  intensity  of  the  tension  or  compression,  =  — -t  ; 

<7,  "  "  "       shear  =  -5-4  . 


PRINCIPAL  AND   CONJUGATE   STRESSES.  247 


and 


10.  Conjugate  Stresses. — Consider  the  equilibrium  of  an 
indefinitely  small  parallelepiped 
abed  (Fig.  216)  of  a  strained  body, 
the  faces  ab,  cd  being  parallel  to 
the  plane  XOX,  and  the  faces  ad, 
be  to  the  plane  YOY. 

Let  the  stresses  on  ab,  cd  act 
parallel   to  the  plane   YOY.     The 
total   stresses    on   ab   and   cd  are 
equal  in  amount,  act  at  the  centres  of  the  faces,  are  parallel  to 
YOY,  and  therefore  neutralize  one  another. 

Hence  the  total  stresses  on  ad  and  be  must  also  neutralize 
one  another.  But  they  are  equal  in  amount,  and  act  at  the 
middle  points  of  ad,  be;  they  must  therefore  be  parallel  to 
XOX. 

Hence,  if  two  planes  traverse  a  point  in  a  strained  body, 
and  if  the  stress  on  one  of  the  planes  is  parallel  to  the  other 
plane,  then  the  stress  on  the  latter  is  parallel  to  the  first 
plane. 

Such  planes  are  called  planes  of  conjugate  stress,  and  the 
stresses  themselves  are  called  conjugate  stresses. 

Principal  stresses  are  of  course  conjugate  stresses  as  well. 

Conjugate  stresses  have  equal  obliquities,  each  obliquity 
being  the  complement  of  the  same  angle. 

n.    Relations     between     Principal     and     Conjugate 

Stresses  (Fig.    217). — Take  any  line  ON  = 


248  THEORY  OF  STRUCTURES. 

P   —  P 

With  N  as  centre  and  a  radius  =  — ,  describe  a  semi- 
circle. 

Let  0  be  the  common  obliquity  of  a  pair  of  conjugate 
stresses. 


FIG.  217. 

Draw  ORS,  making  an  angle  0  with  ON,  and  cutting  the 
semicircle  in  the  points  R  and  5. 
Join  NR,  NS. 

OR  and  OS  are  evidently  a  pair  of  conjugate  stresses. 
Draw  NV  perpendicular  to  RS  and  bisecting  it  in  V. 
Draw  the  tangent  OT\  join  NT. 
Let  OR  =  r,  OS  =  s.    Then 


rs  =  OR.  OS  =  OT*  = 


•and 

=(A+A)cos  0.       (29) 


The  maximum  value  of  the  obliquity,  i.e.,  of  0,  is  the  angle 
TON. 

Call  this  angle  0.     Then 


NT      A -A  /    x 

'  '  '  '  '  (30) 


PRINCIPAL   AND   CONJUGATE   STRESSES. 


249 


Let  OR,  OR'  be  a  pair  of  con- 
jugate stresses  (Fig.  218). 

Let    OG,    OH  be    the    axes  of   R; 
greatest  and  least  principal  stress, 
respectively. 

Draw  ON  normal  to  OR. 

Let  the  angle  GOR  =  #,  RON 
=  6,  HON  =  GOR'  =  7,  as  before.    Then 


0 
FIG.  218. 


and  by  eqs.  (8), 


~  cot  y  =  tan  f/>  =cot 


cot 


•    ,*_  /i  ~A  _  cot  x  —  cot  (y  +  ^)  sin  6 

~  ~  ~:    sin  2 


or 


Hence, 

angle  GON  —  90°  —  y  =  ^ 
and 

angle  HOR  =  y  +  &     =  ±{0  + 


(32) 


.  .     .     (33) 


250  THEORY   OF   STRUCTURES. 

12.  Ratio  of  r  to  s. 


r_OR_  OV-RV  _  OV-)/NR*  -NV* 
s  ~"  OS  *  OV+RV~  OV+VNR*-NVn- 


ONcos  6  -  VNR*  —  ON*  sin3  0 

ON  cos  e  +  VNR*  -  ON*  sin2  e 


But 

NR      A  -  A      NT 


=  sin  TON  =  sin  0. 


r       cos  8  —  1/sin3  <p  —  sin2 


s        cos  6*  -{-  I/sin2  0  —  sin3  6 


cos  <9  —  1/cos2  6*  —  cos2  0 

— ,   .     ,     .     (34) 
cos  (9  +  Vcos2  6^  -  cos2  0 


Let 2  =  sin  a»    Then 

cos  ^ 


r       i  q:  cos  a  a  a 

-  =  —7- =  tan3-    or     =cot2-.  (35) 

j        I  ±  cos  a  2  2 

If  8  =  o,  or  =  90  —  0. 

.-.  j  =  tan2  (45  -  f )     or     =  cot2  (45  -  f  )•    •     (36) 
If  6  =  0,  «  =  90°. 

' (37) 


RELATION  BETWEEN  STRESS  AND   STRAIN.  2$l 

13.  Relation  between  Stress  and  Strain. — Let  a  solid 
body  be  strained  uniformly,  i.e.,  in  such  a  manner  that  lines  of 
particles  which  are  parallel  in  the  free  state  remain  parallel  in 
the  strained  state,  their  lengths  being  altered  in  a  given  ratio, 
which  is  practically  very  small.  Lines  of  particles  which  are 
oblique  to  each  other  in  the  free  state  are  generally  inclined  at 
different  angles  in  the  strained  state,  and  their  lengths  are 
altered  in  different  ratios. 

Let  the  straining  of  the  body  convert  a  rectangular  portion 
ABCD  (Fig.  219)  into  the  rectangle  AB'C'D',  where  AB'  = 
(i  +  a)AB  and  AD'  =  (i  +  ft)AD. 

Now  a  and  ft  are  very  small,  so  that  their  joint  effect  may 
be  considered  to  be  equal  to  the  sum  of  their  separate  effects. 
Hence: 

First.  Let  a  simple  longitudinal  strain  in  a  direction  paral- 
lel to  AB  convert  the  rectangle  ABCD 

into  the  rectangle  AB'ED,  where  BB'  Di _££!_£ 

=  a.AB. 

A  line  OF  will  move  into  the  posi- 
tion OF',  where  FF'  =  a .  DF,  and 

OF'  -OF      O 
the  strain  along  OF  = -=-= 

Mr* 


OF 
8  being  the  angle  OFD. 

Also,  the  "distortion  or  deviation  from  rect  angularity" 


FF'sinB          . 
=  angle  FOF'  =  --  7j/T~  QF  --  ~  a  COS     sm 

Second.  Let  a  simple  longitudinal  strain  in  a  direction 
parallel  to  AD  convert  the  rectangle  ABCD  into  the  rectangle 
ABKD',  where  DD'  —  ft.  AD. 

The  line  OF  will  move  into  the  portions  O'Fn  ',  where 
00'  =  ft.AO  and  F"F=DD'  =  ft.  AD. 

/-\f  Tfli    _ 

.'.  the  strain  along  OF  — 


Ur 


2$2  THEORY  OF  STRUCTURES. 

Draw  O'M  parallel  to  OF.     Then 

O'F"-  OF=  O'F"-  0'M=FffMsm  0  =  (F"F-FM)  sin 
=  (DDr  —  OO')  sin  0  =  ft(AD  -  AQ)  sin  0 
=  ft.  O0  sin  0. 


.          sin  0 
/.  the  strain  along  OF  =  -  --  -  =  ft  sin2  0. 


The  distortion  =  the  angle  F"O'M 
F"Mcos  6      ft.  OP  cos 


OF  OF 


=  ft  sin  0  cos  0. 


Hence,  when  the  strains  are  simultaneous,  the  line  OF  will 
take  the  position  O'F'"  between  O'F"  and  OF',  and 

the  total  strain  along  OF  =  a  cos2  0  +  ft  sin2  0  ; 
the  total  distortion  =  (a  —  ft)  sin  6  cos  0. 

Again,  draw  a  line  OG  perpendicular  to  OF. 

The  angle  OGA  =  90°  —  0,  and  hence,  from  the  above, 

tlae  total  strain  along  OG  =  «  sin2  #  -f-  ft  cos8  0, 
and  the  corresponding  distortion  =  (a  —  ft)  sin  0  cos  0. 


Denote  the  strain  along  <2F  by  *,  ,  that  along  OG  by  ^2  ,  and 
h  of  the  equal  distortions  by  /.    Then 


Again,  if  0/%  <9£  are  the  sides  of  a  rectangle  enclosed  in 
the  rectangle  ABCD,  the  straining  will  convert  the  rectangle 
into  an  oblique  figure  with  its  opposite  sides  parallel.  The 
lengths  of  adjacent  sides  are  altered  by  the  amounts  el  and  e^  , 
and  the  angle  0  by  2t.  The  above  results  may  also  be  consid- 
ered to  hold  true  if  the  straining,  instead  of  being  uniform, 
varies  continuously  from  point  to  point. 


RELATION   BETWEEN  STRESS  AND    STRAIN.  253 

Consider  a  unit  cube  ABCD  subject  to  stresses  of  intensity 
pl   and  /„   upon   the   parallel   faces 
AD,  BC  and  AB,  DC.     By  Art.  3, 
Chap.  Ill, 

_A  _  _A_ 
"  E      mE' 

A    ,  A 


and  the  strain  perpendicular  to  the 


face  ABCD  =  -  -       - 


^?_ 
^£' 

If  the  stresses  are  of  equal  intensity  but  of  opposite  kind, 
i.e.,  if  the  one  is  a  tension  and  the  other  a  compression, 

pl  =  — A  =/,  suppose. 

.*.  a  —  —  /?  =  7-, ( i  +  — ),  and  the  third  strain  is  nil. 
g\     '    mJ 

Thus  the  volume  of  the  strained  solid 

=  (i  4-  a)(i  —  a)(i)  =  i  —  «2  =  i,  approximately, 

so  that  the  volume  is  not  sensibly  changed. 

Also,  if  OGHF  is  an  enclosed  square,  O  being  the  middle 
point  of  AD,  6  =  45°,  and 


el  =  e^  =  =  o  =  strain  along  OF  or  OG, 


and  the  distortion  ~  change  in  angle  O 


a  — 


254  THEORY  OF  STRUCTURES. 

This  result  may  be  at  once  deduced  from  the  figure.     For 

FOG      OD      i  +  ft      i  -  a  790°  -  2A 

tan =  -==  =  — —-  =  — ; —  =  tan  I  - , 

2  FD       i  +  a       l  +  a  \       2       f 

or 

i  —  a  _  i  —  tan  t  _i  —  t 

i  +  a~  l  +  tan  /  ~~  i  +t* 

since  /  is  very  small.     Hence 

t  =  a. 

As  already  shown  in  Art.  3,  shearing  cannot  take  place 
along  one  plane  only,  and  at  any  point  of  a  strained  solid  the 
shears  along  planes  at  right  angles  are  of  equal  intensity.  The 
effect  of  such  stresses  is  merely  to  produce  a  distortion  of 
figure,  and  generally  without  sensible  change  of  volume. 

Thus,  shears  of  intensity  s  along  the  parallel  faces  of  the 
unit  square  ABCD  will  merely  distort  the  square  into  a  rhom- 
bus ABC'D'  (Fig.  221).  Denoting  the  change  of  angle  by  2?, 
and  assuming  that  the  "stress  is  propor- 
tional to  the  strain," 

5  —  G.2t, 

where  G  is  a  coefficient  called  the  modulus 
of  transverse  elasticity,  or  the  coefficient  of 
A      -<— g—    ~~B  rigidity,  and    depends    upon   a  change  of 

FIG.  221.  form. 

Consider  a  section  along  the  diagonal  BD. 

The  stresses  on  the  faces  AB,  AD,  and  on  CB,  CD,  resolved 

parallel  and  perpendicular  to  BD,  are  evidently  equivalent  to 

nil  and  a  normal  force  s  1/2,  respectively.     Thus,  there  is  no 

sliding  tendency  along  BD,  but  the  two  portions  ABD  and 

s  \i  *) 
CBD  exert  upon  each  other  a  pull,  or  tension,  of  intensity  -^- 


_ 

V2~'' 

Similarly  it  maybe  shown  that  there  is  no  tendency  to  slide 
along  AC,  but   that  the   two  portions  ABC  and  ADC  exert 


RANKINGS  EARTHWORK   THEORY.  2$$ 

upon  each  other  a  pressure  of  intensity  s.  The  straining  due 
to  the  shearing  stresses  is,  therefore,  identical  with  that  pro- 
duced by  a  thrust  and  tension  of  equal  intensity  upon  planes 
at  45°.  Hence,  as  proved  above, 


and 


_E     m 


2  I        m      2t 


Now  m  rarely  exceeds  4,  and  hence  G  is  generally  <  \E. 
Again,  the  coefficient  of  elasticity  of  volume,  or  cubic  elasticity 
(Art.  23),  is 


mE  2  (m 

=  -\- 


$(m  —  2)      3  \m  — 
and  hence 

6K+2G 


m  = 


14.  Rankine's  Earthwork  Theory. — A  mass  of  earth- 
work tends  to  take  a  definite  slope. 

Rankine  assumes,  (i)  that  the  stresses  exerted  in  different 
directions  through  a  particle  of  a  granular  mass  are  subject 
to  the  general  principles  enunciated  in  the  preceding  articles  ; 
(2)  that  the  cohesion  of  the  particles  is  gradually  destroyed, 
and  that  the  stability  of  the  mass  ultimately  depends  on  friction 
only. 

In  the  limit,  therefore,  the  face  of  the  mass  is  inclined  to 
the  horizon  at  an  angle  equal  to  the  angle  of  friction,  or,  as  it 
is  sometimes  called,  the  angle  of  repose. 

Adopting  for  the  present  Rankine's  assumptions,  the  equi- 
librium of  the  mass  requires  that  the  direction  of  the  mutual 
pressure  between  the  two  parts  into  which  the  mass  is  divided 
by  a  plane  shall  make  an  angle  with  the  normal  to  the  plane 
less  than  the  angle  of  friction. 


256  THEORY  OF  STRUCTURES. 

Denote  the  angle  of  friction  by  0. 
The  maximum  obliquity  must  be  ^  0. 
By  eq.  (30), 


or  sin 


/,ON 


Thus,  if  a  pressure  of  intensity  /,  acts  through  a  mass  of 
earthwork,  eq.  (38)  gives  the  least  intensity  of  pressure/,  acting 
in  a  direction  perpendicular  to  that  of  /,  consistent  with  equi- 
librium; 

The  limiting  ratios  of  a  pair  of  conjugate  stresses  in  a  mass 
of  earthwork  may  also  easily  be  determined. 

By  eq.  (34), 


cos  0  —  Vcos2  0  —  cos2  0 

the  ratio  = f  ....     (39) 

cos  6+  I/cos2  0-  cos2  0 

Hence  the  ratio  cannot  exceed 


cos  0  -f  ^cos2  6  —  cos2  0 
cos  6  —  Vcos2  0  —  cos2  0 
nor  can  it  be  less  than 


cos  &  —  VCOS*  B  —  cos2  0 


cos  0  +  I/cos2  0  —  cos2  0 
If  6  =  o,  the  ratio  becomes 

I  =f  sin  0 


sin  0 


(40) 


For  example,  let  the  ground-surface  be  horizontal. 
The  pair  of  conjugate  stresses  become  a  vertical  stress 
and  a  horizontal  stress  /, . 

.   A  <  i  +  sin0 


RANKINGS  EARTHWORK    THEORY.  2$? 

or 

A  >  i  -  sin  0 
A  =i+sin0'* 
as  in  eq.  (38). 

Pressure  against  a  Vertical  Plane. — Let  ACB  (Fig.  222), 
the  ground-surface  of  a  mass  of  earthwork,  be  inclined  to  the 
horizon  at  an  angle  6. 

Consider  a  particle  at  a  vertical  depth  CD  =  x  below  C. 

Let  s  be  the  vertical  intensity  of  pressure  on  the  particle 
at  D. 

Let  r  be  the  conjugate  intensity  of  pressure  on  the  particle 
at  D. 

This  conjugate  pressure  acts  in  the  direction  ED  parallel  ta 
the  ground-surface,  and  its  obliquity  is  B. 

Take  DE  so  that 


_r_       cos  0  —  1/cos'  8  —  cos*  0 
DC  '~  s   •"  cos  e  _|_  |/cos*  e  —  cos2  0* 

Then  ze/ .  .£"/?  represents  in  direction  and 
magnitude  the  intensity  of  pressure  on  the 
vertical  plane  DC  at  the  point  D,  w  being  the  weight  of  a  unit 
of  volume  of  the  earthwork. 

Join  CE. 

The  intensity  of  pressure  at    any  other   point  m   is    evi- 
dently w .  mn,  mn  being  drawn  parallel  to  DE. 

Hence,  the  total  pressure   on    the   plane  DC  =  weight    of 
prism  DCE 

w.DC.DE  w.DC*r 

— cos  6  = cos  0 

2  2          S 


_  wx*  cos  6  —  I/cos2  6  —  cos3  0 

~2~  cos  6  +  Vcos2  0  -  cos2  0 '  '     *     (43' 

Again,  s  is  the  pressure  due  to  the  weight  of  the  vertical 
column  CD. 

.-.  s  =  wx  cos  0,      p     ......     (44) 


THEORY  OF  STRUCTURES. 

and 

cos  6  —  t/cos'  6  —  cos2  0 

r  =.  wx  cos  u .     (45) 

cos  6  +  Vcos*  V  -  cos2  0 

"By  means  of  this  last  equation  the  total  pressure  on   CD 
imay  be  easily  deduced  as  follows  : 

The  pressure  on  an  element  dx  at  a  depth  x 


^cos  0  —  I/cos'  0  —  cos'  0  7 
=  77MT  =  wx  cos  0  -  dx. 

cos  0  +  l/cos'J  0  —  cos2  0 

.'.  total  pressure  =   jrdx  =  etc. 

The  total  resultant  pressure  is  parallel  in  direction  to  the 
ground-surface,  and  its  point  of  application  is  evidently  at 
two  thirds  of  the  total  depth  CD. 

15.  Earth  Foundations.  —  CASE  I.  Let  the  weight  of  the 
superstructure  be  uniformly  distributed  over  the  base,  and  let 
pQ  be  the  intensity  of  the  pressure  produced  by  it. 

If  ph  is  the  maximum  horizontal  intensity  of  pressure  cor- 
responding to/0, 

/„_  <  I  +sin  0 
ph  —  i  —  sin  0  ' 

In  the  natural  ground,  let  pv  be  the  maximum  vertical  in- 
tensity of  pressure  corresponding  to  the  horizontal  intensity 

ph  •    Then 

p^  <  i  +  sin  0 

pv  -i  —  sin  0  ' 
Hence 

+sin0\2 


pi  =  VI  —sin  0 

If  ^  is  the  depth  of  the  foundation,  and  w  the  weight  of  a 
cubic  foot  of  the  earth, 

pv  =  wx  ; 
A 


(46) 
wx  =  \i  —  sin  0 


EA  R  TH  FO  UNDA  TIONS.  259 

Let  h  +  x  be  the  height  of  the  superstructure,  and  let  a 
cubic  foot  of  it  weigh  wr.     Then 


Hence,  a  minimum  value  of  x  is  given  by 


x)  _  /i_+j| 
"  -~~  V7-si 


=  p,  suppose; 


wx     '       \i  — sin  0/  ~     /&*' 

»';!# 

•••*=^_^7F2 (47) 

CASE  II.  Let  the  superstructure  produce  on  the  base  a 
uniformly  varying  pressure  of  maximum  intensity /:  and  mini- 
mum intensity/.,. 

By  Case  I, 


Wx  =i-  sn 


(     . 


In  the  natural  ground  the  minimum  horizontal  intensity  of 
pressure  is 

I  —  sin  0 

ph  —  wx  —  ;  —  -  —  -  . 
I  +  sin  0 

When  the  foundation-trench  is  excavated,  this  pressure 
tends  to  raise  the  bottom  and  push  in  the  sides.  The  weight 
of  the  superstructure  should  therefore  be  at  least  equal  to  the 
weight  of  the  material  excavated  in  order  to  develop  a  hori- 
zontal pressure  of  an  intensity  equal 

A  <  *  —  sin 


'  /2  =  I  +  sin  0  ' 
Combining  this  with  the  last  equation, 


Combining  (48)  and  (49), 


(Rankine's  Civil  Engineering,  Arts.  237,  239.) 


200 


THEORY  OF  STRUCTURES. 


16.  Retaining-walls.  —Consider  a  portion  ABMN  of   a 
wall  (Fig.  223). 

Let  Wbe  its  weight,  and  let  the  di- 
rection of  W  cut  MN  in  C. 

Let  Pbe  the  resultant  of  the  forces 
externally  applied  to  ABNM  and  tend- 
ing to  overthrow  it.  Let  D  be  its 
point  of  application,  and  let  its  di- 
rection meet  that  of  W  in  E. 

Let  F  be  the  centre  of  pressure  (or 
resistance)  at  the  bed  MN. 

FIG.  223.  Let  O  be  the  middle  point  of  MN. 

Let  MN  —  t,  OF  —  qt,  OC  —  rt,  q  and  r  being  each  less 
than  unity. 

Let  x'  and  y',  respectively,  be  the  horizontal  and  vertical 
co-ordinates  of  D  with  respect  to  F. 

Let  the    inclination    to  the   horizon  of    MN  =  a,  of  P's 
direction  =  ft. 

Conditions  of  Equilibrium.  —  (a)  The   moment  of   P  with 

respect  to  F  ^  the  moment  of  W  with  respect  to  F,  or 


P(y'  cos  ft  —  x'  sin  ft)  <  W(qt  =F  rt)  cos  a  ; 


(51) 


the  upper  or  lower  sign  being  taken  according  as  C  falls  on  the 
left  or  right  of  O. 

In  ordinary  practice  q  varies  from  J  to  f  . 

EXAMPLE.  —  A  masonry  wall  (Fig.  224) 
of  rectangular  section,  x  ft.  high,  4  ft.  wide, 
weighing  125  Ibs.  per  cubic  foot,  is  built 
upon  a  horizontal  base  and  retains  water 
(weighing  62^  Ibs.  per  cubic  foot)  on  one 
side  level  with  the  top  of  the  wall. 


FIG.  224. 


=  125  X 


=  4  ft. 


RE  TA  IN  ING-  WALLS.  261 

or 

*?  <  192? (52) 

If     0=J,     *2<48     and     x  <  6.928  ft. 
If     0  =  f,     *'  <  72     and     *  <  8.485  ft. 

(£)  The  maximum  intensity  of  pressure  at  the  bed  MN  must 
not  exceed  the  safe  working  resistance  of  the  material  to 
crushing.  The  load  upon  the  bed  is  rarely  if  ever  uniformly 
distributed.  It  is  practically  sufficient  to  assume  that  the  in- 
tensity of  the  pressure  diminishes  at  a  uniform  rate  from  the 
most  compressed  edge  inwards. 

Let  /  be  the  maximum  intensity  of  pressure,  and  R  the 
total  pressure  on  the  bed. 

Three  cases^may  be  considered. 

CASE  I.  Let  the  intensity  of  the  pressure  diminish  uniformly 
from  /  at  M  to  o  at  N  (Fig.  225). 

Take  MG  perpendicular  to  MN  and  =f',  join  GN. 

The  pressure  upon  the  bed    Gp^ 
is  represented  by  the  triangle  ^^ 

MGN.  "\^ 

^- 


MFC 
The  brdinate    through   the  FlG-  22s- 

centre  of  gravity  of  the  triangle,  parallel  to  GM,  cuts  MN  in 
the  centre  of  pressure  F. 

*-  -  -  =  g*. 

CASE  II.    Let  the  maximum  intensity  />  MG  in  Case  I. 

Take  MH  =.  f,  and  the  triangle 

MHK  =  R  (Fig-  226)- 

The   pressure   on  the  bed  is 

^  now  represented  by  the  triangle 

MHK. 

R  =    MH.  MK  =        .  MK. 


M          F  O  K  N 

FlG-  22fi-  The    ordinate    through     the 


262  THEORY  OF  STRUCTURES. 

centre  of  gravity  of  the  triangle  MHK  parallel  to  HM  cuts  MN 
in  the  centre  of  pressure  F. 


.-.  qt  =  OF  =  OM-  MF=--  —. 


But 


t         2  R 
/.  qt  =  ---  -  ;    and  hence 

T  O      C'  T 

9  =  ~  ~  -ft  and  is  evidently  >  g.     .  Y   (53) 

CASE  III.  Let  the  maximum  intensity/  <  MG  in  Case  I. 
Take  ML  =f,  and  the  trapezoid  MLSN  =  R  (Fig.  227). 

The  pressure  on  the  bed  is  now  represented  by  the  trape- 
zoid MLSN. 


G  .-.  R  =  XML  +  NS)MN 


r^. 


and 


MFC  N  2R 

FIG.  «7.  ^V5  =   -  —  /. 

The  ordinate  through  the  centre  of  gravity  of  the  trapezoid 
parallel  to  LM  cuts  MN  in  the  centre  of  pressure  F. 
Draw  ST  parallel  to  NM. 
The  moment  of  MLSN  with  respect  to  O 

=  moment  of  MTSN  with  respect  to  O 
+  moment  of  ZSiTwith  respect  to  (9, 
or 

TS* 

\(ML  +  NS)MN.  OF  = 


Hence,         q  =     (       —  t),  and  is  evidently  <%....     (54) 


RE  TAINING-  WALLS.  263 

Now  W  must  be  a  function  of  x,  the  vertical  depth  of  N 
below  B  ;  P  also  may  be  a  function  of  x. 

Hence  if  /is  given,  and  the  corresponding  value  of  q  from 
(53)  or  (54)  substituted  in  (52),  x  may  be  found. 

When  (53)  is  employed,  the  value  of  x  found  must  make 

?>> 

When  (54)  is  employed,  the  value  of  x  found  must  make 

r<> 

EXAMPLE.  The  rectangular  wall  in  («),  the  safe  crushing 
strength  of  the  material  being  10,000  Ibs.  per  square  foot  (=/). 

R=  W=  500*. 
By  (53), 


~  2         I2O 

Substituting  in  (52), 


Hence, 

x  <  9-03  ft. 

Again,  q  >  --  2j  —  >  .4248,  and  is  a  fortiori  >  g. 

If  (54)  is  employed, 

1/80         \ 
*  =  5b—  V- 
Hence,  by  (52), 


By  trial  ;tr  is  found  to  lie  between  12  and  13  ;  each  of  these 
values  makes  q  >  -J-,  which  is  contrary  to  (54). 

The  first  is  therefore  the  correct  substitution. 

(c\  The  angle  between  the  directions  of  the  resultant  pres- 
sure and  a  normal  to  the  bed  must  be  less  than  the  angle  of 
friction. 


264 


THEORY  OF  STRUCTURES. 


Let  0  be  the  angle  of  friction,  R  the  mutual  normal  pressure. 
Resolving  along  the  bed  and  perpendicular  to  it, 


and 


Pcos  a  4-  ft  —  Wsm  a  <  R  tan  0 

Psin  a  +  J3+  Wcosa  =  R', 
Pcos  a-\-  ft  —  IV sin  a 


Psin 
which  reduces  to 


Wcosa 


<  tan  0, 


P(cos  ft-\-a  cos  0— sin /?-)-«  sin  0)>  W^sin  0  cos  a-j-cos  0  sin  <*), 
or 


P  cos  /?  -f  or  +  0  <  JFsin  a  +  0, 


or 


or 


P(cos  ft  cos  a  -(-  0  —  -  sin  ft  sin  or  +  0)  < 


a  -f-  0, 


tan 


Pcos 


Psin/*  + 


•    •    .    •    (55) 


17.  Rankine's  Theory  of  Earthwork  applied  to  Retain- 
c  ing-walls. — Fig.  228  represents  a 
vertical  section  of  a  wall  retaining 
earthwork.  AB  is  a  vertical  plane 
cutting  the  ground-surface  AC  in  the 
point  A. 

Consider  the  equilibrium  of  the 
whole  mass  of  masonry  and  earth- 
work in  front  of  AB. 

Let  the  depth  AB  —  x. 

The  total  pressure  on  AB  is,  by 
(43), 

_       wx*  cos  6  —  1/cos2  6  —  cos2  0 

P  =  cos  0  -     . 

2  cos  0  +  I/cos3  0  —  cos2  0 


LINE    OF  RUPTURE.  265 

Its  point  of  application  is  D,  and  BD  =  —  . 

Let  W  be  the  weight  of  the  whole  mass  under  considera- 
tion, and  let  its  direction  cut  the  base  of  the  wall  in  the  point  G. 
Let  F  be  the  centre  of  pressure  in  the  wall-base. 
Taking  moments  of  jPand  W7  about  F, 


p    -  cos  6  —  j[jf  <f  j)  sin(0+a)  |  =   W(qt  ^  rt)  cos  a.  (56) 


The  other  conditions  of  equilibrium  may  be  discussed  as  in 
Art.  1 6. 

18.  Line  of  Rupture. — Another  expression   for  the  press- 
ure on  AB  may  be  obtained  as  follows  : 

If  the  whole  mass  in  front  of  AB  (Fig.  229)  were  suddenly 
removed,  some  of  the  earthwork  behind  AB  would  fall  away. 

Suppose  that  the  volume  ABC  would  slip 
along  the  plane  CB. 

The  stability  of  ABC  is  maintained  by  the 
reaction  P  on  AB,  the  weight  J^of  ABC,  and 
the  frictional  resistance  along  BC. 

Let  the  direction  of  P  make  an  angle  ft 
with  the  horizon. 

Let  the  angle  CBA  =  i. 

Let  R  be  the  mutual  pressure  on  the  plane  BC. 

Resolving  along  and  perpendicular  to  BC, 

-  P  cos  (90°  -  i  -  /?)  +  Wcos  i  =  R  tan  0 ; 
and 

P  sin  (90°  -i- ft)+  W sin  i  =  R. 

.'.  -  Psin  (ft+t)  +  W cos  i  =  tan  0  \Pcos  (fl+  J)+  Wsin  i j, 
and 

cos  i  —  sin  t  tan  0  cos(/+0) 

=  ^  sin  (ft  +  i}  +  cos  (ft  +  i)  tan  0  ~     Y  sin  (0  +  i  +  0)' 


266  THEORY  OF  STRUCTURES. 

Also 


BA  .  BC  .    .      wx*  cos  0  sin  i 

W  —  w sin  i  — 

2  2    cos 


_  w^r2  cos  0  sin  z      cos  (t  +  0) 

*' 


2    cos  (0+)  sh(/J  +  *  +  0) 


The  only  variable  upon  which  P  depends  is  the  angle  i. 

Differentiating  the  right-hand  side  of  eq.  57  with  respect  to 
i  and  putting  the  result  equal  to  zero,  a  value  of  i  is  found  in 
terms  of  /?,  0  and  0,  which  will  make  P  a  maximum. 

The  line  inclined  at  this  angle  to  the  vertical  is  called  the 
line  of  rupture. 

If  the  ground-surface  is  horizontal,  9  =  o. 

If  the  face  retaining  the  earth  is  vertical,  and  if  it  is  also 
assumed  that  the  friction  between  the  face  and  the  earthwork  is 
nil,  P  is  horizontal  and  ft  =  o.  Hence  (57)  becomes 


P  =  —  tan  i  cot  (i  +  0) (58) 

This  is  a  maximum  when  21  =  90°  —  0,  and  then 


/i  - 

WX*  f  0\  /  0\  WX1! 

=  —  tan  (45  -  -)  cot  (45  +  2-J=  —  (    • 

\  J    * 


or 

__  wx*  i  —  sin  0 
2    I  +  sin  0  ' 

the  same  result  as  that  obtained  by  Rankine's  theory. 


PRACTICAL   RULES. 


267 


The  following  is  an  easy  geometrical  proof  of  eq.  (59): 

On  any  line  KL  (Fig.  230)  de- 
scribe a  semicircle. 

Draw  KM  inclined  at  the  angle 
0  to  KL,  and  KN  inclined  at  the 
angle  i  to  KM.  . 

Join  NL,  cutting  KM  in  T. 

Let  O  be  the  middle  point  of 
the  arc  KM. 

Join  OL,  cutting  KM  in  Y. 

Draw  NV  parallel  to  KM.    Then 

NT     KN     NT      VY 


FIG.  230. 


VY 
The  ratio  -==     is  evidently  a  maximum  when  N  coincides 


with  0,  and  hence  tan  i  cot  (i  +  0)  is  a  maximum  when 
coincides  with  KO. 

Now  the  arc  OK  =  the  arc  6W,  and  hence  the  angle  OKM 
=  the  angle  OLK. 

Hence,  if  OKM=  i,  OLK  must  also  =  i. 

But  OKL  +  OLK=  90°  =  *'+  0  +  *  =  2*'+  0. 


19.  Practical  Rules.  —  When  the  surface  of  the  earthwork 
is  horizontal  and  the  face  of  the  wall  against  which  it  abuts  ver- 
tical, the  pressure  on  the  wall  according  to  Rankine's  theory  is 


sin  0 


2    i  +  sin  0  ' 


and  the  direction  of  P  is  horizontal. 

This  result  is  also  identical  with  that  obtained  in  Art.  18,  on, 
the  assumption  of  Coulomb's  wedge  of  maximum  pressure 
(Poncelet's  Theory). 


268  THEORY  OF  STRUCTURES. 

Experience  has  conclusively  proved  that  the  theoretical  value 
of  P  given  above  is  very  much  greater  than  its  real  value,  so 
that  the  thickness  of  a  wall  designed  in  accordance  with  theory 
would  be  in  excess  of  what  is  required  in  practice.  In  the 
deduction  of  the  formula,  indeed,  the  altogether  inadmissible 
assumption  is  made  that  there  is  no  friction  between  the  earth- 
work and  the  face  of  the  wall.  This  is  equivalent  to  the  sup- 
position that  the  face  is  perfectly  smooth  and  that  therefore 
the  pressure  acts  normally  to  it.  Boussinesque,  Levy,  and  St. 
Venant  have  demonstrated  that  the  hypothesis  of  a  normal 
pressure  only  holds  true, 

either,  first,  if  the  ground  surface  is  horizontal  and  the  wall- 

0 

face  inclined  at  an  angle  of  45° to  the  vertical, 

or,  second,  if  the  wall-face  is  vertical  and  the  ground-surface 
inclined  at  an  angle  0  to  the  horizon. 

When  the  surface  of  the  ground  is  horizontal  and  the  face 
of  the  wall  vertical,  and  when  0  =  45°,  the  above  formula  gives 
the  correct  magnitude  of  P.  Its  direction,  however,  is  not  hori- 
zontal, but  makes  an  angle  with  the  vertical  equal  to  the  angle 
of  friction  between  the  earth  and  the  wall.  The  wall-face  is  gen- 
erally sufficiently  rough  to  hold  fast  a  layer  of  earth,  and  in  all 
probability  Boussinesque's  assumption  that  the  friction  between 
the  wall  and  the  earth  is  equal  to  that  inherent  in  the  earth  is 
a  near  approximation  to  the  truth.  The  direction  of  P  will 
thus  be  considerably  modified,  leading  to  a  smaller  moment  of 
stability  and  a  corresponding  diminution  in  the  necessary  thick- 
ness of  the  wall. 

In  practice  the  thrust  P  may  always  be  made  small  by 
carrying  up  the  backing  in  well-punned  horizontal  layers. 

In  order  to  neutralize  the  very  great  thrust  often  induced 
by  alternate  freezing  and  thawing  and  the  consequent  swelling, 
a  most  effective  expedient  is  to  give  a  batter  of  about  I  in  I  to 
the  rear  line  of  the  wall  extending  below  the  line  to  which  frost 
penetrates. 

The  greatest  difficulty  in  formulating  a  table  of  earth-thrusts 
arises  from  the  fact  that  there  is  an  infinite  variety  of  earth- 
work. As  an  example  of  this,  Airy  states  that  he  has  found 


PRACTICAL   RULES.  269 

the  cohesive  power  of  clay  to  vary  from  168  to  800  pounds  per 
square  foot,  the  corresponding  coefficients  of  friction  varying 
from  1.15  to  .36,  and  that  even  this  wide  range  is  less  than 
might  be  found  in  practice. 

A  correct  theory  for  the  design  of  retain  ing-walls  is  as  yet 
wanting.  According  to  Baker,  experience  has  shown  that  with 
good  backing  and  a  good  foundation  the  stability  of  a  wall 
will  be  insured  by  making  its  thickness  one-fourth  the  height, 
and  giving  it  a  front  batter  of  I  or  2  in.  per  foot,  and  that  under 
no  conditions  of  ordinary  surcharge  or  heavy  backing  need  its 
thickness  exceed  one-half  the  height.  Baker's  usual  practice  in 
ground  of  average  character  is  to  make  the  thickness  one-third 
the  height  from  the  top  of  the  footings,  and  if  any  material  is 
taken  out  to  form  a  face  panel,  three-fourths  of  it  is  put  back 
in  the  form  of  a  pilaster. 

General  Fanshawe's  rule  for  brick  walls  of  rectangular 
section  retaining  ordinary  material  is  to  make  the  thickness 

24$  of  the  height  for  a  batter  of     in    5  ; 


or"       « 


2/7  « 


«  tt  tt  «          « 


«          « 


Oo  «       it  tt  t(  tt         tt 

~  «  «  II  tt  tl  « 


in  6  ; 
in  8  ; 
in  10 ; 


in  12 ; 
in  24; 
32 "     "  "  "         "       for  a  vertical  wall. 

The  thickness  at  the  footing  adopted  by  Vauban  for  walls 
with  a  front  batter  of  I  in  5  or  I  in  6  and  plumb  at  the  rear,  is 
approximately  given  by  the  empirical  formula 

thickness  =  .  19/^  +  4  ft., 

H  being  the  height  of  the  wall  above  the  footing.  Counter- 
forts were  introduced  at  intervals  of  15  feet  for  walls  above  35 
feet  in  height,  and  at  intervals  of  12  feet  for  walls  of  less  height. 

TT 

The  counterfort  projects  from  the  wall  a  distance  of (-3  ft. 

approximately,  and  the  approximate  width  of  the  counterfort  is 

TT  TT 

—  -f-  3  ft.,  diminishing  to  —  +  2  ft. 


2/0  THEORY  OF   STRUCTURES. 

Brunei  curved  the  face  of  the  wall  and  made  its  thickness 
one-fifth  or  one-sixth  the  height.  Counterforts  2  ft.  6  in.  in 
thickness  were  introduced  at  intervals  of  10  ft. 

The  vast  importance  of  the  foundation  will  be  better  appre- 
ciated by  bearing  in  mind  that  the  great  majority  of  failures 
have  been  due  to  defective  foundations.  If  water  can  percolate 
to  the  foundation,  a  softening  action  begins  and  a  consequent 
settlement  takes  place,  which  is  most  rapid  in  the  region  sub- 
jected to  the  greatest  pressure,  viz.,  the  toe.  In  order  to  coun- 
teract this  tendency  to  settle,  the  toe  may  be  supported  by  rak- 
ing piles,  the  rake  being  given  to  diminish  the  bending  action  of 
the  thrust  on  the  piles.  It  is  also  advisable  to  distribute  the 
weight  as  uniformly  as  possible  over  the  base,  a  condition  which 
is  not  compatible  with  large  front  batters  and  deep  offsets,  as 
they  tend  to  concentrate  weight  on  isolated  points.  In  the 
case  of  dock-walls,  too,  a  large  front  batter  will  keep  a  ship 
farther  away  from  the  coping  and  will  necessitate  thicker 
fenders,  as  well  as  cranes  with  wider  throws.  As  an  objection 
to  offsets  Bernays  urges  that,  in  settling,  the  backing  is  liable 
to  hang  upon  them,  forming  large  holes  underneath.  He  there- 
fore favors  the  substitution  of  a  batter  for  the  offsets.  On  the 
other  hand,  if  water  stands  on  both  sides  of  the  walls,  the 
hydrostatic  pressure  on  the  offsets  will  greatly  increase  its 
stability. 

Dock-walls  are  liable  to  far  greater  variations  of  thrust  than 
ordinary  retaining-walls.  The  water  in  a  dock  with  an  im- 
permeable bottom  may  stand  at  a  much  higher  level  than  the 
water  at  the  back  of  the  wall,  and  its  pressure  may  thus  even 
more  than  neutralize  the  thrust  due  to  the  backing.  With  a 
porous  bottom  the  stability  of  a  wall  may  be  greatly  dimin- 
ished by  an  upward  pressure  on  the  base.  The  experience  of 
dock-wall  failures  has  led  to  the  conclusion  that  a  large  moment 
of  stability  is  not  of  so  much  importance  as  "  weight  with  a 
good  grip  on  the  ground."  Many  authorities,  both  practical 
and  theoretical,  have  urged  the  great  advantages  in  economy 
and  strength  attending  the  employment  of  counterforts.  The 
use  of  Portland  cement,  or  cement  concrete,  will  guard  against 
the  breaking  away  of  the  counterforts  from  the  main  body  of 


RESERVOIR    WALLS. 


271 


the  wall,  as  has  often  happened  in  the  case  of  the  older  walls. 
But  a  uniform  distribution  of  pressure  as  well  as  of  weight  is 
important,  and  it  therefore  seems  more  desirable  to  introduce 
the  extra  weight  of  the  counterforts  into  the  main  wall.  Be- 
sides, the  building  of  the  counterforts  entails  of  itself  an  in- 
creased expense. 

20.  Reservoir  Walls. — Let  /"be  the  maximum  safe  press- 
ure per  square  foot  of  horizontal  base,  at  inner  face  of  a  full 
reservoir,  at  outer  face  when  empty. 

Let  w  be  the  weight  of  a  cubic  foot  of  the  masonry. 

Assume  that  the  wall  is  to  be  of  uniform  strength,  i.e.,  that 
the  section  of  the  wall  is  of  such  form  that 
in  passing  from  any  horizontal  section  to  the 
consecutive  one  below,  the  ratio  of  the  in- 
crement of  the  weight  to  the  increment  of 
the  surface  is  constant  and  equal  to/. 

Let  AB,  Fig.  231,  be  the  top  of  the  wall. 
Take  any  point  O  as  origin,  and  the  vertical 
through  O  as  the  axis  of  x. 

Let    OA  —  t,,    O£=r /2,  and  let 
T  =  /,  4-  tn  -- 


B  0 


For  the  profile  AP  consider  a  layer  of  thickness  dx  at  a 
depth  x.     Then 

wydx 


(0 


or 


w  y 

,^/-lc 

w 

c  being  a  constant  of  integration. 

When  x  =  o,  y  =  tl ; 


•'•  °  =  -  lo&  'i  +  '» 


2/2  THEORY  OF  STRUCTURES. 

and  hence 

*  =    *" 


which  is  the  equation  to  AP  and  is  the  logarithmic  curve. 
It  may  be  similarly  shown  that  the  equation  to  BQ  is 


Equations  (2)  and  (3)  may  also  be  written  in  the  forms 

V) 

y  =  t,efx    .....  ,>,  ,    .     .    (4) 


and 


(5) 


Corresponding  points  on  the  profiles,  e.g.,  P  and  <2,  have  a 
common  subtangent  of  the  constant  value  —  ,  for 


=      .....     (6) 
dy)       w 


Area  PNOA  =  ydx  =  f  .  -  =  (  F,  -  ,,),  (7) 
where  /W=  F,. 

Area  QNOB  =  f'yd»  =  £-(  Y,  -  t,),  (8) 
where  QN  =  Y,. 


.:  Area  QPAB  =    -(  K,+  F,  -  «,  +  ^)  =     (T'  -  7"),   .    (9) 
where  PQ=Y,+  Y,=  T'. 


RESERVOIR    WALLS. 

Thus  the  area  of  the  portion  under  consideration  is  equal 
to  the  product  of  the  subtangent  and  the  difference  of  thick- 
ness at  top  and  bottom. 

Lines  of  resistance  with  reservoir  empty.  Let  gl  be  the 
point  in  which  the  vertical  through  the  C.  of  G.  of  the  portion 
OAPN  intersects  PN.  Then 


Ngl  X  area  OAPN  =   f*Xydxy-  ; 

I/O 


So  if  gi  be  the  point  in  which  the  vertical  through  the 
C.  of  G.  of  the  portion  OBQN  intersects  QN, 


Let  G  be  the  point   in  which  the  vertical    through   the 
C.  of  G.  of  the  whole  mass  ABQP  intersects  PQ.     Then 

NG  X  area  ABQP  =  Ng,  X  area  A  ONP-  Ng,  X  area  BONQ, 


or 


IV  A.  W  l  A.  W  * 


274 


THEORY  OF  STRUCTURES. 


The  horizontal  distance  between  G  and  a  vertical  through 
the  middle  point  of  AB 

=NG-\(f  ^jy-'x-w-wjy-^-w*-** 

•=.  one  half  of  the  horizontal  distance  between  the  verticals 
through  the  middle  points  of  AB  and  CD. 

The  locus  of  G  can  therefore  be  easily  plotted. 

Lines  of  Resistance  with  Reservoir  Full. — Let  R  be  the  centre 
of  resistance  in  PQ  (Fig.  232). 

Draw  the  vertical  QS,  and  consider 
the  equilibrium  of  the  mass  QSAPQ. 

Let  w'  =  weight  of  a  cubic  foot  of 
water. 

w'x*  x 

=  moment  of  water-pressure 

against  QS  about  R 
=  moment  of  weight  of  QBS 
about   R  -f-   moment    of 
weight  of  QPAB  about  Rt 
or 


W  X 


—  moment  of  QBS  about  R+  ~(T'  -  T)w.GR. 


The  first  term  on  the  right-hand  side  of  this  equation  is 
generally  very  small  and  may  be  'disregarded,  the  error  being 
on  the  safe  side. 
In  such  case 

i  w'       x* 
GR-6jT'-T' 

Also  the  mean  intensity  of  the  vertical  pressure 

w  X  area  APQB 
—  A  = or> — 


RESERVOIR    WALLS. 

and  the  maximum  intensity  of  the  vertical  pressure 


275 


2R 


=  */ 


I  -2q 


or 


=  -Jv(l  +  6?)  =  /(!  +  6?)(l  -  y,). 

General  Case. — Let  the  profile  be  of  any  form,  and  consider 
any  portion  ABQP,  Fig.  233. 

Take  the  vertical  through  Q  as  T 
the  axis  of  x,  and  the  horizontal 
line  coincident  with  top  of  wall  as 
the  axis  of  y. 

The  horizontal  distance  (y)  be- 
tween the  axis  of  x  and  the  vertical 
through  the  C.  of  G.  of  the  portion 
under  consideration  is  given  by  the 
equation 


/  being  the  width,  dx  the  thickness, 

and  y  the  horizontal  distance  from  OQ  of  the  C.  of  G.  of  any 

layer  MN  at  a  depth  x  from  the  top. 

When  the  reservoir  is  empty,  the  deviation  of  the  centre  of 
resistance  from  the  centre  of  base 


When  the  reservoir  is  full,  let  q'T  be  the  deviation  of  the 
centre  of  resistance  from  the  centre  of  the  base,  and  disregard 
the  moment  of  the  weight  of  the  water  between  OQ  and  the 
profile  BQ.  Then 


276 


THEORY  OF  STRUCTURES. 


moment  of  water  pr.  ±  moment  of  wt.  of  ABQP 
weight  of  ABQP 


w  I    tdx 

t/o 


Hence 


w'x* 


w  I    tdx 

t/o 


21.  General  Equations  of  Stress.  —  Let  x,  y,  z  be  the 
co-ordinates  with  respect  to  three  rectangular  axes  of  any  point 

O  in  a  strained  body. 

Consider  the  equilibrium  of  an 
element  of  the  body  in  the  form 
of  an  indefinitely  small  parallele- 
piped with  its  edges  OA  (=  dx\ 
OB(—  dy),  OC(—  dz]  parallel  to 
the  axes  of  x,  y,  z.  It  is  assumed 
that  the  faces  of  the  element  are 
sufficiently  small  to  allow  of  the 
distribution  of  stress  over  them 
being  regarded  as  uniform.  The 
FlG-  234'  resultant  force  on  each  face  will 

therefore  be  a  single  force  acting  at  its  middle  point. 
Let  Xl  ,  Yl  ,  Z^  be  the  components  parallel  to  the  axes  x,  y,  z 
of  the  resultant  force  per  unit  area,  on  the 
face  EC. 
"    X^  ,  Y9  ,  Z9  be    the    corresponding    components    for   the 

taceAC. 
"    -Yg  ,  Y3,  Z3  be    the    corresponding    components    for   the 

face  AB. 

These  components  are  functions  of  x,  y,  z,  and  therefore 
become 


for  the  adjacent  face  AD  ; 


GENERAL  EQUATIONS  OF  STRESS. 


for  the  adjacent  face  BD  ; 


for  the  adjacent  face  DC. 
Hence,  the  total  stress  parallel  to  the  axis  of  x 

=  X.dydz  -  (x,  +  d-^dx]dydz  +  X,dzdx  -  (x,  +  ^ 


IdX,  .  dX, 

=  ~(-&+W 

Similarly,  the  total  stress  parallel  to  the  axis  of  y 


and  the  total  stress  parallel  to  the  axis  of  z 

t      dZt      dZ 


Let  p  be  the  density  of  the  mass  at  O,  and  let  Px,Py,  Pg  be 
the  components  parallel  to  the  axes  of  x,  y,  z  of  the  external 
force,  per  unit  mass,  at  O. 

pdxdydzPx  is  the  component  parallel  to  the  axis  of  x  of 

the  external  force  on  the  element  ; 
pdxdydzPy  is  the  component  parallel  to  the  axis  of  y  of 

the  external  force  on  the  element  ; 

pdxdydzPz  is  the  component  parallel  to  the  axis  of  z  of 
the  external  force  on  the  element. 


278  THEORY  OF  STRUCTURES. 

The  element  is  in  equilibrium. 


dx    l     dy         dz 
dY,   ,  dY,   ,   dY, 


dZ \    ,dZ^  >dZ* 
dx"  dy"  dz~ 


These  are  the  general  equations  of  stress. 

Again,  take  moments  about  axes  through  the  centre  of  the 
element  parallel  to  the  axes  of  co-ordinates,  and  neglect  terms 
involving  (dxjdydz,  dx(dyfdz,  dxdy(dz}*. 


and    X= 


(2) 


Adopting  Lamp's  notation,  i.e.,  taking 
Nl ,  Nt  ,  N3  as  the  normal  intensities  of  stress  at  O  on  planes 

perpendicular  to  the  axes  of  x,  y,  z  ; 
7",  as  the  tangential  intensity  of  stress  at  O  on  a  plane 
perpendicular  to  the  axis  of  x  if  due  to  a  stress 
parallel  to  the  axis  of  j/,  or  on  a  plane  perpen- 
dicular to  the  axis  of  y  if  due  to  a  stress  parallel 
to  the  axis  of  x  ;  and  7^ ,  Ta  similarly, — equa- 
tions (i)  become 


^  +  ~^'~Jr~^1 
rf'+tL+W 

dx        dy        dz 


(3) 


GENERAL  EQUATIONS   OF   STRESS. 


279 


Next    consider  the  equilibrium  of   a  tetrahedral   element 
having  three  of  its  faces  parallel  to 
the  co-ordinate  planes.     Let  /,  m,  n 
be  the  direction-cosines  of  the  normal 
to  the  fourth  face. 

Also,  let  X,  V,  Z  be  the  compo- 
nents parallel  to  the  axes  of  x,  y,  z 
of  the  intensity  of  stress  R  on  the 
fourth  face. 

X  =  IN,  +  mT,  +  n  Tt  +  \pPxldx. 

But  the  last  term  disappears    in  FlG-  235- 

the  limit  when  the  tetrahedron  is  indefinitely  small,  and  hence 


(4) 


These  three  equations  define  R  in  direction  and  magnitude 
when  the  stresses  on  the  three  rectangular  planes  are  known. 

Let  it  be  required  to  determine  the  planes  upon  which  the 
stress  is  wholly  normal.  We  have 


Y=mR,    Z—nR.. 


(5) 


Substituting  these  values  of  X,  Y,  Z  in  eqs.  (4)  and  eliminat 
ing  /,  m,  n,  we  obtain 


-  T?-  T?-  T? 
1T,TS)  =  o  ;   (6) 


R*  - 


a  cubic  equation  giving  three  real  values  for  R,  and  therefore 
three  sets  of  values  for  /,  m,  and  n,  showing  that  there  are  three 
planes  at  O  on  each  of  which  the  intensity  of  stress  is  wholly 
normal.  These  planes  are  at  right  angles  to  each  other  and 
are  called  principal  planes,  the  corresponding  stresses  being  prin- 
cipal stresses.  They  are  the  principal  planes  of  the  quadric, 


=  c. 


(7) 


280 


THEORY  OF  STRUCTURES. 


For,  the  equation  to  the  tangent  plane  at  the  extremity  of  a 
radius  r  whose  direction-cosines  are  /,  m,  n  is 

Xrx  +  Yry  +  Zrz  =  c,      ....'.     (8) 
and  the  equation  of  the  parallel  diametral  plane  is 

Xx  +  Yy  +  Zz  =  o •     •    (9) 

The  direction-cosines  of  the  perpendicular  to  this  plane  are 


X 
- 


Y 

- 


Z 

- 


so  that  the  resultant  stress  R  must  act  in  the  direction  of  this 
perpendicular. 

Hence  the  intensities  of  stress  on  the  planes  perpendicular 
to  the  axes  of  the  quadric  (7)  are  wholly  normal. 

Refer  the  quadfic  to  its  principal  planes  as  planes  of  refer-* 
ence.  All  the  7"s  vanish  and  its  equation  becomes 


(10) 


Also,  the  general  equations  (3)  become 


%-*••• 


Again, 


RELATION  BETWEEN  STRESS  AND   STRAIN.          28 1 

Consider  X9  Y,  Z  as  the  co-ordinates  of  the  extremity  of 
the  straight  line  representing  R  in  direction  and  magnitude. 
Equation  (12)  is  then  the  equation  to  an  ellipsoid  whose  semi- 
axes  are  Nlt  N91  Nt.  As  a  plane  at  O  turns  round  O  as  a  fixed 
centre,  the  extremity  of  a  line  representing  the  intensity  of 
stress  R  on  the  plane  will  trace  out  an  ellipsoid.  This  ellipsoid 
is  called  the  ellipsoid  of  stress. 

Note  I.  The  coefficients  in  the  cubic  equation  (6)  are  in- 
variants. Thus,  NI  -f  N9  +  N,  is  constant,  or  the  sum  of  three 
normal  intensities  of  stress  on  three  planes  placed  at  right 
angles  at  any  point  of  a  strained  body  is  the  same  for  all 
positions  of  the  three  planes. 

Note  2.  The  perpendicular/  from  O  on  the  tangent  plane, 
equation  (8), 


Note  3.  Let  the  stress  be  the  same  for  all  positions  of  the 
plane  at  0.  Then  N,  =  N^  =  Nz ,  and  the  ellipsoid  (12)  be- 
comes a  sphere.  The  stress  is  therefore  everywhere  normal, 
and  the  body  must  be  a  perfect  fluid.  Conversely,  if  the 
stress  is  everywhere  normal,  the  body  must  be  a  perfect  fluid, 
the  ellipsoid  becomes  a  sphere,  and  therefore  Nl=  N9  =  Na. 

22.  Relation  between  Stress  and  Strain. — In  Art.  13  it 
was  shown  that  when  the  size  and  figure  of  a  body  are  altered 
in  two  dimensions,  there  is  an  ellipse  of  strain  analogous  to  the 
ellipse  of  stress.  If  the  alteration  takes  place  in  three  dimen- 
sions, it  may  be  similarly  shown  that  every  state  of  strain  may 
be  represented  by  an  ellipsoid  of  strain  analogous  to  the  ellip- 
soid of  stress.  The  axes  of  the  ellipsoid  are  the  principal  axes 
of  strain,  and  every  strain  may  be  resolved  into  three  simple 
strains  parallel  to  these  axes. 


282 


THEORY  OF  STRUCTURES. 


It  is  assumed  that  the  strains  remain  very  small,  that  the 
stresses  developed  are  proportional 
to    the    corresponding    strains,    and 
that  their  effects  may  be  superposed. 
Consider  an   element  of  the  un- 
strained  body  in   the  form  of  a  rect- 
x  angular    parallelepiped,    having    its 
edges  PQ  (=  A),  PR  (=  k},  PS  (=  I) 
parallel  to  the  axes  of  co-ordinates. 

When   the  body  is  strained,  the 
element  becomes  distorted,  the  new 
edges  being  P'Q',  P'R',  P'S'. 

Let  x,  y,  z  be  the  co-ordinates  of  P. 

Let  x  -f-  u,  y  -f-  v,  z  -\-  w  be  the  co-ordinates  of  P'. 

By  Taylor's  Theorem  the  co-ordinates  with  respect  to  P'  of 


FlG>  236' 


du 
— 
dx 


dv  ,dw 
*-—,  h— 
dx  dx 


r>  ,         i.du       J     ,   dv  \       ,dw 
R    are  k—,     k(  i  +-=-)»     k—  ; 
a  \        a   i        a 


du       .dv       .       .   dw\ 
— 


r,         ,  . 

S'are/-—,     /—-  , 
dz        dz 


» 
ay  i        ay 

dw\ 
—  . 

dz  j 


Hence,  strain  parallel  to  axis  of  x  = 

«  <fc  («  y       _ 


P'Q'-PQ    du  1 
PQ  ~  ~dx 

P'R'-PR    dv 


z  = 


P'S' -PS    dw 

PS       ~~dz' 


I  SO  TROPIC  BODIES. 

Again,  cos  QP'R 


283 


du 


dv  \dv        dw  dw 


In  the  limit,  this  reduces  to 


Similarly, 


dy       dx 
cos  fi'P'S'  =      -  + 


...    (16) 


cos  *'/"<?'  = 


Volume  of  unstrained  element  =  hkl; 


Volume  of  distorted  element  = 


Difference  of  volume 


multiplied  by  the  cosines 
of  small  angles 

,   du    .  dv    .   dw\ 

+  _  +  _.+_), 

in  the  limit. 


du    .   dv    .   dw 


'  Vol.  of  unstrained  element       dx       dy       dz* 

=  the  volume  or  cubic  strain. 

23.  Isotropic  Bodies,  i.e.,  bodies  possessing  the  same  elas- 
tic properties  in  all  directions. 

A  normal  stress  of  intensity  Nl  parallel  to  the  axis  of  x 

N 

produces   a   simple   longitudinal   strain  —-±,  and   two   simple 

E 

N 

lateral  strains,  each  = -r,  parallel  to  the  axes  of  y  and  z, 

mh 


284 


THEORY  OF   STRUCTURES. 


E  being  the  ordinary  modulus  of  elasticity  and  —  ,  Poisson's 


ratio  (Art.  3,  Chap.  III). 

Normal  stresses  N9  ,  Nt  parallel  to  the  axes  of  y  and  a  may 
be  similarly  treated. 

Let  the  three  normal  stresses  act  simultaneously  and  super- 
pose the  results.  Then 


,        ,a    du  ~\ 
total  strain  parallel  to  axis  of  x  —  -—  l  --  —  —  ?=  —  • 

E         mE        dx  ' 


mE        dz  ' 


The   form  in  which  these  equations  are  given  is  due  to 
Grashof. 

Solving  for  Nt  ,  N^  ,  N9  , 


,,          m(m  —  i)E      du  mE          tdv       dw\ 

1  ~  (m  +  i)(m—  2)  dx  "•  (m  +  i)(m-  2)<dy    '   ~dzr 

„          m(m  —  I  )E     dv  mE           idw  .du\ 

~  ~~ 


., 
~~ 


m(m  —  i  }E     dw 


mE 


(m  +  i)(;«—  2)       ~"  (w  +  i)(m—  2)\dx 
The  last  equations  may  be  written 


.dv 


(du    >dv\ 
I  7Z~   i   "717  /  * 


. 
dy 


ISOTROPIC  BODIES. 


285 


where  A,  = 


mE 


,  is  the  coefficient  of  dilatation,  and 


A  =      mm~l 

~  (m+i)(m-2J 

Again,  the  straining  changes  the  angle  RPS  by  an  amount 

—  -| — - ,  producing  two  tangential  stresses,  each   equal   to 
dy        dz 

G\  —  -| — -  I,  parallel  to  the  axes  of  y  and  z. 

\dv        dz> 


dw       dv 


Similarly, 


T  -G(— 

2"      \dz 
du 


s- 


(21) 


G  is  called  the  coefficient  of  rigidity  or  transverse  elasticity, 
and  is  designated  n  in  Thomson  and  Tait's  notation,  and  >w  in 
Lamp's  notation. 

Relation  between  A,  A,  and  G. — Equations  (20)  and  (21)  pre- 
serve the  same  forms  whatever  rectangular  axes  may  be  chosen. 

Keep  the  axis  of  z  fixed  and  turn  the  axes  of  x  and  y 
through  an  angle  a. 

Let  Nf  be  the  normal  stress  parallel  to  the  new  axis  of  x. 

.*.  N{  —  N^  cos2  a  -|-  N^  sin2  a  +  2  7"3  sin  a  cos  a.       (22) 

Let  x1 ',  y'  and  z/,  v'  be  the  new  co-ordinates  and  displace- 
ments. 

,T.        Adu'    .    ^Idv'    .   dw'\        ,.       ..du'    . 

•'•  N>  =  A^'  + A (^  +  w)  = (A  ~  ^ + K6- 

_      du   .   dv   .   dw  du'    .   dv'    .   dw'  . 

For  --  +  --  +  -:-,  =  0  =  y-,  +  — •  +  -— - ,  is  an  invariant. 
dx       dy       dz  dx        dy        dz 


.286  THEORY  OF  STRUCTURES. 

The   values   of  N{  given  by  eqs.  (22)  and  (23)  must  be 
identical.     Now, 

x  =  x'  cos  a  —  y'  sin  a,    y  =  x'  sin  a  -)-  y'  cos  a  ;      } 
z/'  =  &  cos  a  -f-  ^  sin  a',     t;'  =  —  &  sin  a  -\-  v  cos  a  .  ) 

du        du  .   dv 

.'.  _  =  —  cos  a  +  —  sin  a 
^,r        dx  dx 

du        2       .   dv    .  a  /^«    .   ^/z/\  . 

=  -j-  cos2  a  +  —  -  sin  a  +  !_--[-__  jsm  a  cos  a 
^r  dy  \dy        dxl 

du  .   dv  T. 

—-  —  cos2  a  -f-  —  -  sin3  a  +  —  ?  sin  a  cos  a  ; 
tfJtr  «y  6- 

and  by  eq.  (23), 

AV  =  (A  -*)(^  cos8  a+^  sin3  a+5  sin  a  cos  a]  +  A0.     (25) 
>«^r  ay  G-  ' 

Also  by  eqs.  (20)  and  (22), 


=  (A—\)-  cos2  «+-  sin  «  +r  sin  a  cos  a+A0.  (26) 
\d>x  dy  A  ~~  A  / 

Eqs.  (25)  and  (26)  must  be  identical. 
^  _  A  w.fi' 

IT  =  5(^+1)  =  "  =  n-  |iK  *  (27) 

Adding  together  equations  (20), 


/    ,   flfe    ,  d 
^"dy^  ~dz' 

It  may  be  easily  shown  that  the  normal  stresses  can  each 
be  separated  into  a  fluid  pressure  p  and  a  distorting  stress. 


AP PLICA  TIONS.  287 

Hence,  putting 

AT  AT  AT__^_  m&  (d11       I      dV      .     6 

9-     3  —  P  --  ^m  __  ^  [^  -r  j-  -r  - 

t>  mE 

the  cubic  elasticity  —  -.—  — =-  =  —. r  =  K.  (28) 

au        dv        aw        ^(m  —  2) 

dx       dy       dz 

24.  Applications.  — i.  Traction. — One  end  of  a  cylindrical 
bar  of  isotropic  material  is  fixed  and  the  bar  is  stretched  in  the 
direction  of  its  length.  The  axis  of  the  bar  is  the  only  line 
not  moved  laterally  by  contraction. 

Take  this  line  as  the  axis  of  x. 

The  displacements  u,  v,  w  of  any  point  x,  y,  z  may  be  ex- 
pressed in  the  form 

u  =  ax,     v  =  —  Pyt     w  =  —  PZ.    .     .     ,     (29) 
By  eqs.  (20)  and  (29), 

.- 1         (30) 


By  eqs.  (21)  and  (29),  all  the  tangential  stresses  vanish. 

Hence,  since  Nlt  N9,  N3  are  constant,  and  since  the  equa- 
tions of  internal  equilibrium  contain  only  differential  coeffi- 
cients of  the  stresses,  the  hypothesis,  eq.  (29),  satisfies  these 
equations. 

First.  Let  N^  =  o  =  N3 ;  i.e.,  let  no  external  force  act  upon 
the  curved  surface. 

.-.  -  ft  A  +  \(ft  +  «)  =  o, 
or 

P  _        A       _  I 

a  ~  T+Ji  ~  m 

Thus,  the  coefficient  of  contraction  is  less  than  the  coefficient 
of  expansion. 


288  THEORY  OF  STRUCTURES. 

Again,  by  eqs.  (30)  and  (32), 

Z  =  A-2*£  =  A-*  =  £.  (33) 

a.  am 

Second.  If  the  bar,  instead  of  being  free  to  move  laterally, 
has  its  surface  acted  upon  by  a  uniform  pressure  P,  then 

N,  =  N3  =  P. 
By  eqs.  (3 1)  and  (32), 

ft  AP—XN, 


a  ~  \(N,  +  2/>)  -  AN, ' 


•-   *    •    .    (35) 


For  example,  let  P  be  sufficient  to  prevent  lateral  contrac- 
tion.    Then  ft  =  o  and,  by  eqs.  (31)  and  (35), 


aA  =  N,  =  -y  =  (m  -  i)P. 

2.  Torsion. — (a)  Let  a  circular  cylinder  (hollow  or  solid)  of 
length  /  undergo  torsion  around  its  axis  (the  axis  of  x),  and  let 
t  be  the  angle  through  which  one  end  is  twisted  relatively  to 
the  other.  A  point  in  a  transverse  section  distant  x  from  the 

latter  will  be  twisted  through  an  angle  x— . 

The  displacements  u,  v,  w  of  the  point  x,y,  z  in  this  section 
may  be  expressed  in  the  form 

u  =  o,     v  =  —  zx-y     w  =  -\-yx- . 
By  eqs.  (20)  and  (21), 

and 


APPLICA  TIONS.  289 

The  algebraic  sum  of  the  moments  of  7!, ,  J",  with  respect 
to  the  axis 


r  being  the  distance  of  the  point  (x,  y,  z)  from  the  axis. 

Hence,   the  moment  M,  =  Pp  (Chap.  IX),  of   the   couple 
producing  torsion 

_    ^  *     /      i  J  c*  __    /~*      T /~*f}T 

~lJ    '  1 

dS  being  an  element  of  the  area  at  (x,y,  z\  I  the  polar  moment 
of  inertia,  and  0  the  torsion  per  unit  of  length  of  the  cylinder, 
or  the  rate  of  twist. 

The  torsional  rigidity  of  a  solid  cylinder 

=  —  =  GI  =  -nR* 

6  2 

R  being  the  radius  of  the  cylinder. 

(b]  Torsion  of  a  bar  of  elliptic  section. 

The  displacements  u,  v,  w  may  now  be  expressed  in  the 
form 


u  =  F(y,  z),     v  =  —  6x2,     w  =  Qxy. 

du  _  dv  _  dw^ 

'  dx  dy        dz' 


7-0,     T.=  G+ 
Hence,  by  the  general  eqs.  (3), 

*  .......  (36) 


2QO  THEORY  OF   STRUCTURES. 

.Also,  the  surface  stresses  are  zero ; 


-and  hence,  by  eqs.  '(3$), 


—  ~dy  =  Q(zdz  -\-ydy) (38) 


This  equation  must  hold  true  at  the  surface. 
Let  the  equation  to  the  elliptic  section  be 


dz  fy 

=  -•     •     •    .^v,,,  •    (40) 


and  by  eq.  (38), 

2     dll         .         „     dlt      _     _    0         /  72    _       2X 

u  —  dyz  satisfies  this  last  equation  and  also  eq.  (36),  if 


Again,  the  algebraic  sum  of  the  moments  71,  ,  T3  with  re- 
spect to  the  axis  of  x, 


2G6 
T(<y  +  *v)  ........    (43) 


APPLICATIONS.  291 

The  total  moment  (M]  of  the  couple  producing  torsion 


and  the  torsional  rigidity 


(c]  Torsion  of  a  bar  of  rectangular  section. 
As  in  case  (b),  u  must  satisfy  the  equation 


Also,  the  equations  of  condition  corresponding  to  eq.  (38)  are 


and 


du 

-j-  —  02  =  o    when    y  =  ±  #,  .     .     .     .     (46) 


—  +  By  =  o     when     z  =  ±  c\       ...     (47) 


2b  and  2c  (b  <  c)  being  the  sides  of  the  rectangle.     The  total 
moment  of  torsion,  viz.,/(7'2j  —  T.z)dS  is  then  found  to  be 


,.     , 

-' 


tan  h   2n       i 


If  ^  =  r,  i.e.,  if  the  section  is  a  square,  eq.  (48)  becomes 

M=  .843462/£0,     ......     (49) 


/(  =  f  ^4)  being  the  moment  of  inertia  with  respect  to  the  axes.. 
(See  Chap.  IX). 


2Q2  THEORY  OF  STRUCTURES. 

If  —  is  very  small,  eq.  (48)  becomes 


M=  i6FcGO(--.2i-) (50) 

The  torsional  rig'dity  of  a  rectangular  section  is  sometimes  ex- 
pressed by  the  formula 

M        5      £V 

T>*4*q^ <5I> 

For  the  further  treatment  of  this  subject,  the  student  is  re- 
ferred to  St.  Venant's  edition  of  Clebsch,  and  to  Thomson  and 
Tait's  Natural  Philosophy. 

3.  Work  done  in  the  small  strain  of  a  body  (Clapeyron's 
Theorem)  — M  ultiply  eqs.  (3)  \yyudxdy  dz,  v  dx  dy  dz,  wdxdydz, 
and  find  the  triple  integral  of  their  sum  throughout  the  whole 
of  the  solid. 

The  terms  involving  the  components  Px  t  Py  ,  P2  may  be  dis- 
regarded, as  the  deformations  due  to  their  action  are  generally 
inappreciable. 

Also, 

///s-***  ; 

=ff(Nx'ux  -  Nx"ux"}dydz  -fffNSj-dxdydz\ 

Nx',  Nx"  being  the  values  of  Nl  at  the  two  points  in  which  the 
line  parallel  to  the  axis  of  x  cuts  the  surface  of  the  body,  and 
ux  i  ux"  the  corresponding  values  of  u. 

Let  dS,  dS'  be  the  elementary  areas  of  the  surface  at  these 
points,  and  /',  I"  the  cosines  of  the  angles  between  the  normals 
to  these  elements  and  the  axis  of  x. 

The  double  integral  on  the  right-hand  side  of  the  last  equa- 
tion then  becomes 

ff(Nx'l'ux'dS  -  NJ'l"Hx"dS)  = 


APPLfCA  TfOlfS  293 

Treating  the  other  terms  similarly, 

O  =  2{(Nll+  T,m  +  7»a  +(T,l+  N,m 


Hence,  the  work  done  =  $2(Xu  +  Yv  +  Zw)dS 


-  ^W,  +  N,N,  +  NJtt  -  T?  -  77  -  T.*)  } 


I 


M  +  Njft  +  ffjrt  -  T:  -  T;  -  T,' 


E  being  the  ordinary  modulus  of  elasticity. 


294  THEORY  OF  STRUCTURES. 


EXAMPLES. 

1.  At  a  point  within  a  strained  solid  there  are  two  conjugate  stresses, 
viz.,  a  tension  of  200  Ibs.  and  a  thrust  of  150  Ibs.  per  square  inch,  the 
common  obliquity  being  30°.     Find  (a)  the  principal  stresses  ;  (b)  the 
maximum  shear  and  the  direction  and  magnitude  of  the  corresponding 
resultant  stress;  (c)  the  resultant  stress  upon  a  plane  inclined  at  30°  to 
the  axis  of  greatest  principal  stress. 

Ans. — (a)  A  tension  of  204.65  Ibs.  and  a  thrust  of  146.95  Ibs.  per  sq.  in. 
(b)  175.8  Ibs.  per  sq.  in.;    173.2  Ibs.  in  a  direction  making  an 
angle  of  40°  13'  with  the  axis  of  greatest  principal  stress. 
(<r)  163.3  Ibs.  per  sq.  in. 

2.  A  wall  with  a  plumb  rear  face  is  to  be  30  ft.  high  and  4  ft.  wide  at 
the  top ;   the  earth  slopes  up  from  the  inner  edge  at  the  angle  of  20°, 
30°  being  the  angle  of  repose.     Assuming  Rankine's  theory,  determine 
the  proper  width  of  the  base,  the  masonry  weighing  144  Ibs.  per  cubic 
foot,  and  the  earth  1 10  Ibs. 

3.  A  wall  6  ft.  wide  at  the  bottom,  plumb  at  the  rear,  and  with  a 
front  batter  of  i  in  12,  retains  water  level  with  the  top.      Find  (a)  the 
limiting  position  of  the  centre  of  pressure  at  the  base  so  that  the  stress 
may  be  nowhere  negative. 

How  (b)  high  may  the  wall  be  built  when  subjected  to  this  condition? 
(a  cubic  foot  of  masonry  =  125  Ibs.). 

Ans.  (a}  12  in.  from  middle  point  of  base;  (b)  height  =  8.9  ft. 

4.  A  wall  is  built  up  in  layers,  the  water  face  being  plumb  and  the 
rear  stepped.     If  /  be  the  thickness  of  the  n\.\\  layer  and  y  the  depth  of 
water  above  its  lower  face,  show  that  width  of  layer  x  thickness  of  layer 
=  t/4^42  +  6Ats  +  mty*  —  2A  ;   A  being  the  sectional  area  of  the  wall 
above  the  layer  in  question,  z  the  horizontal  distance  between  the  water 
face  and  the  line  of  action  of  the  resultant  weight  above  the  layer,  t  the 
layer's  thickness,  and  m  the  ratio  of  the  specific  weights  of  the  water  and 
masonry. 

5.  At  a  point  within  a  strained  solid,  the  stresses  on  two  planes  at 
right  angles  to  each  other  are  a  thrust  of  30  ^2  Ibs.  and  a  tension  of  60 
Ibs.  per  square   inch,  the   obliquities   being  45°  and   30°   respectively. 
Determine  (a)  the  principal  stresses;    (b)  the  ellipse  of  stress;   (c)  the 
intensity  of  stress  upon  a  plane  inclined  at  60°  to  the  major  axis. 

Ans. — (a)  A  thrust  of  61.76  Ibs.  and  a  tension  of  39.80  Ibs. 
(c)  A  thrust  of  66.5  Ibs. 


EXAMPLES.  295 

6.  If  the  principles  of  the  ellipse  of  stress  are  applicable  within  a 
mass  of  earth,  and  if  at  any  point  of  the  mass  the  stress  upon  a  plane  is 
double  its  conjugate  stress,  the  angle  between  the  two  stresses  being 
20°  28',  show  that  the  angle  of  repose  of  the  earth  is  28°.  i. 

7.  The  total  stress  at  a  point  O  upon  a  plane  AB  is  60  Ibs.  per  square 
inch,  and  its  obliquity  is  30° ;  the  normal  component  upon  a  plane  CD 
at  the  point  O  is  40  Ibs.  per  square  inch ;  CD  is  perpendicular  to  AB. 
Find  (a)  the  total  stress  upon  CD,  and  also  its  obliquity  ;  (b)  the  princi- 
pal stresses  at  O  ;  (c)  the  equal  conjugate  stresses  at  O. 

Ans. — (a)  tan -'(I) ;  50  Ibs. 

(b)  76.57  Ibs.  and  15.39  Ibs. 

(c)  34.23  Ibs, ;  obliquity  =  41°  42'. 

8.  Assuming  Rankine's  theory,  find  the  pressure  on  the  vertical  face 
of  a  retaining-wall,  30  ft.  high,  which  retains  earth  sloping  up  from  the 
top  at  the  angle  of  repose,  viz.,  30°. 

(Weight  of  masonry  =  128  Ibs.  per  cubic  foot.;  weight  of  earth  =  120 
Ibs.  per  cubic  foot.)  Ans.  46,764  Ibs. 

9.  At  a  point  within  a  strained  solid  the  stress  on  one  plane  is  a  ten- 
sion of  50  Ibs.  per  square  inch  with  an  obliquity  of  30°,  and  upon  a 
second  plane  is  a  compression  of  150  ibs.  per  square  inch  with  an  ob- 
liquity of  45°.     Find  (a)  the  principal  stresses  ;  (ff)  the  angle  between  the 
two  planes  ;  (c}  the  plane  upon  which  the  resultant  stress  is  a  shear,  and 
the  amount  of  the  shear. 

Ans. — (a)  PI  =  1 53. 8  Ibs.  (comp.)  ;  p*  =  —  20  Ibs.  (tens.) 

(*)  21°  55'- 

(c)  86.88  ibs.;  y  =  19°  50'. 

10.  At  a  point  within  a  strained  solid  the  stress  on  one  plane  is  a  ten- 
sion of  loo  Ibs.  per  square  inch  with  an  obliquity  of  30°,  and  on  a  second 
plane  a  compression  of  50  Ibs.  with  an  obliquity  of  60°.     Find  (a)  the 
angle  between  the  planes  ;  (b)  the  plane  upon  which  the  stress  is  wholly 
a  shear;  (c)  the  planes  of  principal  stress. 

Ans.— (a)  11°  38'. 

(b)  64.6  Ibs. ;  y  =  3°  26'. 

(c)  pi  =  106.46  (tens,) ;  p*  =  —  39.26  (compr.). 

11.  In  the  preceding  question  find  the  conjugate  stresses  at  the  given 
point  having  the  common  obliquity  45°.  Ans.  Impossible. 

12.  At  a  point  within  a  strained  mass  the  principal  stresses  at  a  given 
point  are  in  the  ratio  of  3  to  i.     Find  the  ratio  of  the  conjugate  stresses 
at  the  same  point  having  the  common  obliquity  30°.     Also  find  the  in- 
clination of  the  axis  of  greatest  principal  stress  to  the  horizontal. 

Ans.  Equal ;  60°. 

13.  A  wall  3  feet  thick,  of  rectangular  section  and  weighing  125  Ibs. 
per  cubic  foot,  is  subjected  to  a  horizontal  thrust  of  800  Ibs.  per  foot  rum 


296  THEORY  OF  STRUCTURES. 

at  its  top.  What  should  be  the  height  of  the  wall  in  order  that  all  the 
joints  above  the  base  may  be  f nationally  stable  ?  Coefficient  of  friction 
=  unity.  Ans.  12  ft. 

14.  A  wall  12  ft.  high,  2  ft.  wide  at  the  top,  and  3  ft.  wide  at  the  bot- 
tom, is  constructed  of  masonry  weighing  120  Ibs.  per  cubic  foot.     The 
overturning  force  on  the  rear  face  of  the  wall,  which  is  plumb,  is  a  hori- 
zontal force  P  acting  at  4  ft.  from  the  base.     Find  P  so  that  the  devia- 
tion of  the  centre  of  pressure  in  the  base  may  not  exceed  \  ft.     The 
centre  of  pressure  being  fixed  at  2  in.  from  the  middle  of  the  base,  show 
that  |  of  the  section  may  be  removed  without  altering  its  stability,  and 
find  the  increase  in  the  inclination  of  the  resultant  pressure  on  the  base 
to  the  vertical,  consequent  on  the  removal. 

Ans.  360  Ibs.;  tangents  of  angles  are  in  ratio  of  5  to  3. 

15.  A  reservoir  wall  is  4  ft.  wide  at  top,  has  a  front  batter  of  i  in  12,  a 
rear  batter  of  2  in  12,  and  is  constructed  of  masonry  weighing  125  Ibs. 
per  cubic  foot;  the  maximum  compression  is  not  to  exceed  12,800  Ibs. 
per  square  foot.     Find  the  limiting  height  of  the  wall. 

Ans.  24  ft.,  q  being  f  §• 

1 6.  A  dock-wall,  plumb  at  the  rear  and  having  a  face  with  a  batter  of 
i  in  24,  is  20  ft.  high  and  9  ft.  wide  at  the  base.     Counterforts  are  built 
at  intervals  of   12  ft.,  projecting  3  ft.  from  the   rear  and  6  ft.  wide. 
Determine  the  thickness  of  an  equally  strong  wall  without  counterforts, 
with  the  same  face-batter  and  also  plumb  in  the  rear. 

Ans.  10.95  ft- 

17.  If  the  walls  in  the  preceding  question  are  founded  in  earth  weigh- 
ing 112  Ibs.  per   square  foot  and   having  an   angle  of   repose  of  32°, 
find  the  least  depth  of  foundation  in  each  case,  the  masonry  weighing 
125  Ibs.  per  cubic  foot.  Ans.  2.72  ft.  ;  2.71  ft. 

1 8.  A  vertical  retajning-wall  is  strengthened   by  means  of   vertical 
rectangular  anchor-plates  having  their  upper  and  lower  edges  18  and  22 
ft.,  respectively,  below  the  surface.     Find  the  holding  power  per  foot  of 
•width,  the  earth  weighing  130  Ibs.  per  cubic  foot  and  having  an  angle  of 
repose  of  30°.  Ans.  27,733^  Ibs. 

19.  Determine  the  limiting  depths  of  foundation  for  (a)  a  wall  of 
rectangular  section  20  ft.  high  ;  (b}  for  a  wall  of  trapezoidal  section  hav- 
ing plumb  rear  and  front  faces  4  and  20  ft.  high  respectively.     Angle 
of  repose  of  earth  =30°;   weight  of  earth  =  112  Ibs.  per   cubic  foot; 
of  masonry  =  140  Ibs.  Ans.  (a)  3.22  ft. ;  (b)  1.93  ft. 

20.  A  wall  20  ft.  high  and  6  ft.  thick  retains  earth  on  one  side  level 
with  the  top,  and  on  the  other  the  earth  rises  up  the  wall  at  its  natural 
slope,  viz.,  45°,  to  the  height  of  5  ft.     Will  the  wall  stand  or  fall  ? 

(Weight  of  masonry  per  cubic  foot  =  130  Ibs.;  of  earth  =  120  Ibs.) 
Find  the  locus  of  the  centres  of  pressure  of  successive  layers. 


EXAMPLES.  297 

Ans.  Overturning  moment  =  4128  ft.-lbs  ;  moment  of  stability 
=  93600?  +  750  (¥  —  6?)  =  369!  2i  ft.-lbs    if  q  =  f. 
The  wall  is  stable. 

21.  The  upper  half  of  the  section  of  a  masonry  wall  is  a  rectangle 
4  ft.  wide,  and  the  lower  half  a  rectangle  6  ft.  wide,  one  face  being  plumb. 
Find  the  height  of  the  wall  so  that  the  stress  on  the  base  may  nowhere 
exceed  10,000  Ibs.  per  square  foot  when  the  wall  retains  water  (a)  on  the 
plumb  face,  (b)  on  the  stepped  face. 

(Masonry  weighs  125  Ibs.  per  cubic  foot.) 

Ans.  (a)  13.08  ft.;  (b)  9.8  ft. 

22.  A  masonry  dam  h  ft.  high  is  a  right-angled  triangle  ABC  in  sec- 
tion, and  retains  water  on  the  vertical  face  AB.    Show  that  the  thickness 

4/£u 

t  of  the  base  BC  is  given  by  f  =  —  -  -  ,  qt  being  the  deviation  of  the 

0 


centre  of  pressure  in  the  base  from  the  middle  point. 

4/£2 
Also  show  that  the  thickness  will  be  given  by  f  =  ——  -  -  if  the 


rock  upon  which  the  wall  is  built  is  seamy,  and  if  it  is  assumed  that  the 
communication  between  the  water  in  the  seams,  and  that  in  the  reservoir 
produces  an  upward  pressure  upon  the  base  BC,  varying  uniformly  from 
that  equivalent  to  the  head  at  B  to  nil  at  C.  If  q  =  $,  show  that,  in 
order  that  the  wall  may  slide,  the  coefficient  of  friction  must  be  less  than 
67  per  cent  in  the  first  and  81  per  cent  in  the  second  case. 

(Weight  of  a  cubic  foot  of  masonry  =  2£  x  weight  of  cubic  foot  of 
water.) 

23.  A  wall  30  ft.  high  is  of  triangular  section  ABC*  the  face  AB  being 
plumb,  and  water  being  retained  on  the  side  AC  level  with  the  top  of  the 
wall  ;  the  masonry  weighs  125  Ibs.  per  cubic  foot.     Find  the  thickness  of     / 
the  base  BC  (a)  when  q  =  f  ;  (b)  when  stress  in  masonry  is  not  to  exceed 
10,000  Ibs.  per  square  foot  ;  (c)  when  q  =  f  and  the  wall  also  retains  earth 
on  the  side  AB  level  with  the  top,  the  angle  of  repose  being  30°. 

Ans.  (a)  17.69  ft.;  (b)  13.19  ft.  ;  (c)  17  ft. 

24.  A  wall  4  ft.  wide  at  the  top,  with  a  front  batter  of  I  in  8,  and  a 
rear  batter  of  i  in  12,  is  30  ft.  high.     Will  the  wall  be  stable  or  unstable 
(i)  when  it  retains  water  level  with  the  top  ;  (2)  when  it  retains  earth? 

(Weight  of  masonry  per  cubic  foot  =  125  Ibs.;  of  earth  =  112  Ibs.  ; 
angle  of  repose  =  30°  ;  and  q  =  f  .) 

Ans.  (i)  Moment  of  wt.  =  128,863  ft.-lbs.  ;  overturning  moment 

=  281,250  ft.-lbs.,  and  wall  is  therefore  unstable. 
(2)  Moment  of  wt.  =  148,251  Ibs.;   overturning  moment 
=  168,000  Ibs.,  and  wall  is  therefore  unstable. 

25.  The  faces  of  a  reservoir  wall  4  ft.  wide  at  top  and  40  ft.  high  have 
the  same  batter,  and  water  rises  on  one  side  to  within  6  ft.  of  the  top. 
Find  the  batter,  assuming  (a)  that  tbs  pressure  on  the  horizontal  base  is 


298  THEORY  OF  STRUCTURES. 

to  be  nowhere  negative ;  (b)  that  the  pressure  varies  uniformly  and  at 
no  point  exceeds  10,000  Ibs.  per  square  foot. 

(Weight  of  masonry  =125  Ibs.  per  cubic  foot.) 

Ans.  (a)  35.8  ft.;  (b)  30  ft. 

26.  The  faces  AB,  AC  of  a  wall  are  parabolas  of  equal  parameters  hav- 
ing their  vertices  at  B  and  C\  water  rises  on  one  side  to  the  top  of  the 
wall.     Determine  the  thickness  of  the  horizontal  base  BC,  (a)  for  a  wall 
50  ft.  high ;  (b)  for  a  wall   100  ft.  high,  so  that  the  pressure  on  the  base 
may  at  no  point  exceed   10,000  Ibs.  per  square  foot.     Also  (c)  compare 
the  volume  of  such  wall  with  the  volume  of  an  equally  strong  wall  of  the 
same  height,  but  with  a  section  in  the  form  of  an  isosceles  triangle  with 
its  vertex  at  A. 

(Weight  of  masonry  =  125  Ibs.  per  cubic  foot.) 

Ans.   («)  32.44  ft. ;  (b)  119. 17  ft. 

(c}  in  case  (a)  ratio  =  7  :  |/i  18 ; 
"      (b)     "     =4/156:21. 

27.  The  water-face  AC  of  a  wall  has  a  batter  of  i  in  10 ;  the  width  of 
the  wall  AD  at  the  top  is  6  ft. ;  the  rear  of  the  wall  DEF  has  two  slopes, 
DE,  having  a  batter  of  2  in  10,  and  EF,  a  batter  of  78  in  TOO;  the  masonry 
weighs  125  Ibs.  per  cubic  foot,  and  the  maximum  compression  must  not 
exceed  85  Ibs.  per  square  inch.    Find  the  safe  heights  of  the  two  portions 
AE  and  EC. 

28.  The  section  ABCD  of  a  retain  ing-wall  for  a  reservoir  has  a  verti- 
cal face  BC  and  a  parabolic  water-face  AD,  with  the  vertex  at  D.     The 
width  of  the  base  DC  —  4  x  width  of  the  top  AB.     If  AB  =  6  ft.,  find 
the  height  of  the  wall,  and  trace  the  curves  of  resistance  (a)  when  the 
reservoir  is  full ;  (b)  when  empty. 

(Cubic  foot  of  masonry  =  2  x  cubic  foot  of  water.) 
Ans.  32  ft.  if  q  =  \,  and  then  max.  compn.  =  8000  Ibs.  per  sq.  ft. 

29.  The  figure  represents  the  section  of  the  upper  portion  of  a  masonry 

dam  which  has  to  retain  water  level  with  the  top  of 
the  dam.  The  face  AC  is  plumb  for  a  depth  of  73  ft. 
The  width  of  the  section  is  constant  and  =  22-J  ft.  for 
a  depth  AB  =  40  ft. 

Find  the  maximum  stress  in  the  masonry  at  the 


\  horizontal  bed  BF.     With  the  same  maximum  stress, 

\          what  should  be  the  width  of  the  horizontal  bed  CG, 

\        FG  being  straight  ? 
^  ~  Q          (Masonry  weighs  130  Ibs.  per  cubic  foot.) 

FlG-  a37-  Ans.  20,720  Ibs.  per  sq.  ft. 

30.  A  wall  of  an  isosceles  triangular  section  with  a  base  36  ft.  wide 
has  to  retain  water  level  with  its  top.     How  high  may  such  a  wall  be 
built  consistent  with  the  condition  that  the  stress  in  the  masonry  is 
nowhere  to  exceed  10,546!  Ibs.  per  square  foot  ? 
(Weight  of  masonry  per  cubic  foot  =  125  Ibs.) 

Ans.   54  ft.,  and  q  =  &. 


EXAMPLES.  299 

31.  When  a  cylindrical  bar  is  twisted,  show  that  it  is  subjected  to 
shears  along  transverse  and  radial  longitudinal  sections,  or  to  tensions 
and  compressions  on  helices  at  45°  to  the  axis. 

32.  Find  the  work  done  in  gradually  and  uniformly  compressing  a 
body  of  volume  V\  to  the  volume  Fa  ,  p  being  the  final  intensity  of 
pressure  and  k  the  modulus   of  compression.      Also  show   that  the 
intensity  of  stress  is  constant  throughout  the  body. 

An,.    £. 


33.  A  bar  is  stretched  under  a  force  of  intensity/.     If  the  bar  is  pre- 
vented from  contracting,  find  the  lateral  stress  ;  also  find  the  extension. 

An,.    -£-';    t'n"-m-^. 
m  —  I       h.    m(in  —  i) 

34.  Taking  the  value  of  the  coefficient  of  elasticity  (E)  and  the  co- 
efficient of  rigidity  (G)  to  be  15,000  and  5750  tons  for  steel,  13,950  and 
5450  tons  for  wrought-iron,  and  9500  and  3750  tons  for  cast-iron,  find  the 
coefficient  of  elasticity  of  volume  (AT),  and  also  the  values  of  the  direct 
elasticity  (A)  and  the  lateral  elasticity  (/I),  assuming  the  metals  to  be 
isotropic. 

'    m  K  A  A 

Ans.  Steel  ............  3f          1277?! 

Wrought-iron...  3f  4        10559! 

Cast-iron  ........   3!  6785^-  ^-G  \G 

35.  A  body  is  distorted  without   compression  or    expansion  ;    find 
the  work  done. 

Ans.   -1   C\NS  +  AV  +  N3*  +  2  (TV  +  TV  +  T 

36.  Find  the  work  required  to  twist  a  hollow  cylinder  of  external 
radius  7?i  ,  internal  radius  RZ  ,  and  length  /  through  an  angle  a. 

ncr* 

Ans.    n—(RS  —  Rf). 
4/ 

Prove  that  torsion  is  equivalent  to  a  shear  at  each  point. 

37.  Show  that  a  simple  elongation  is  equivalent  to  a  cubical  dilation 
and  a  pair  of  shearing  or  distorting  stresses. 

38.  Find  the  resultant  shearing  stress  at  any  point  in  the  surface  of 
the  transverse  section  of  an  elliptic  cylinder.     (Art.  24,  Case  b.) 

Ans.  26—       c    a  ,  p  being  the  perpendicular  from  the  centre 

upon  the  tangent  to  the  ellipse  at  the  given  point,  and 
2b,  2.c  the  major  and  minor  axes. 

39*.  A  cylinder  undergoes  torsion  round  its  axis.      Show  that  the 
curves  of  no  traction  are  concentric  circles. 


CHAPTER  V. 
FRICTION. 

I.  Sliding  Friction. — Friction  is  the  resistance  to  motion 
which  is  always  developed  when  two  substances,  whether  solid, 
liquid,  or  gaseous,  are  pressed  together  and  are  compelled  to 
move  the  one  over  the  other.  If  P  is  the  mutual  pressure,  and 
if  F  is  the  force  which  must  act  tangentially  at  the  point  of 
contact  to  produce  motion,  the  ratio  of  ^to  Pis  called  the  co- 
efficient of  friction  and  may  be  denoted  by/.  The  value  of/ 
does  not  depend  upon  the  nature  of  any  single  substance,  but 
upon  the  nature  and  condition  of  the  surfaces  of  contact  of  a 
pair  of  substances.  It  is  not  the  same,  e.g.,  for  iron  upon  iron 
as  for  iron  upon  bronze  or  upon  wood  ;  neither  is  it  the  same 
when  the  surfaces  are  dry  as  when  lubricated. 

The  laws  of  friction  as  enunciated  by  Coulomb  are  : 

(i)  That  /  is  independent  of  the  velocity  of  rubbing  ;  (2) 
that /is  independent  of  the  extent  of  surface  in  contact;  (3) 
that /depends  only  on  the  nature  of  the  surfaces  in  contact. 

The  friction  between  two  surfaces  at  rest  is  greater  than 
when  they  are  in  motion,  but  a  slight  vibration  is  often  suffi- 
cient to  change  the  friction  of  rest  to  that  of  motion. 

Morin's  elaborate  friction  experiments  completely  verified 
these  laws  within  certain  limits  of  pressure  (from  f  Ib.  to  128 
Ibs.  per  square  inch)  and  velocity  (the  maximum  velocity  being 
10  ft.  per  second),  and  under  the  conditions  in  which  they 
were  made. 

A  few  of  his  more  important  results  are  given  in  the  follow- 
ing table  : 

300 


SLIDING  FRICTION. 


301 


Material. 

State  of  Surfaces. 

Coefficient  of  Friction. 

Wood  on  wood 

dry.  . 

.25  tO  .5 

Metal  on  wood 

dry.  . 

.2       "  .6 

>  <       «        >« 

wet  

22    "  .26 

Metal  on  metal 

dry  

.15    "  .2 

wet  .  . 

.3 

Metal  and  wood  c 

.15 

on    each    other  1 
or  each  on  itself  ' 

occasionally  lubricated  as  usual.  .  .  . 
constantly  lubricated   

.07  to  .08 

.OS 

The  apparatus  employed  in  carrying  out  these  experiments 
consisted  of  a  box  which  could  be  loaded  at  pleasure,  and 
which  was  made  to  slide  along  a  horizontal  bed  by  means  of  a 
cord  passing  over  a  pulley  and  carrying  a  weight  at  the  end. 
The  contact-surfaces  of  the  bed  and  box  were  formed  of 
the  materials  to  be  experimented  upon.  The  pull  was  meas- 
ured and  recorded  by  a  spring  dynamometer. 

More  recent  experiments,  however,  have  shown  that 
Coulomb's  laws  cannot  be  regarded  as  universally  applicable, 
but  that  /  depends  upon  the  velocity,  the  pressure,  and  the 
temperature.  At  very  low  velocities  Morin's  results  have 
been  verified  (Fleeming  Jenkin).  At  high  velocities  f  rap- 
idly diminishes  as  the  velocity  increases.  Franke,  having 
carefully  examined  the  results  of  various  series  of  experi- 
ments, especially  those  of  Poire"e,  Bochet,  and  Galton,  has 
suggested  the  formula 


v  being  the  velocity  and/0,  a,  coefficients  depending  upon  the 
nature  and  condition  of  the  rubbing  surfaces. 

For  example, 

fn  =  .29  and  a  =  .04  for  cast-iron  on  steel  with  dry  sur- 
faces. 

/0  —  .29  and  a  =  .02  for  wrought-iron  on  wrought-iron  with 
dry  surfaces. 

/0  =  .24  and  a  —  .0285  for  wrought-iron  on  wrought-iron 
with  slightly  damp  surfaces. 

Ball  has  shown  that  at  very  low  pressures  f  increases  as 


3C2  THEORY  OF  STRUCTURES. 

the  pressure  diminishes,  while  Rennie's  experiments  indicate 
that  at  very  high  pressures/  rapidly  increases  with  the  press- 
ure, and  this  is  perhaps  partly  due  to  a  depression,  or  to  an 
abrasion  of  the  rubbing  surfaces. 

2.  Inclined  Plane. — Let  a  body  of  weight  P  slide  uni- 
formly up  an  inclined  plane  under  a  force  Q  inclined  at  an 
angle  ft  to  the  plane. 

Let  F  be  the  friction  resisting  the  mo- 
tion, R  the  pressure  on  the  plane,  and  a  the 
plane's  inclination. 

The  two  equations  of  equilibrium  are 

F  =  Q  cos  j3  —  P  sin  a 
and 

R  =  —  Q  sin  ft  +  P  cos  ex. 

F          Qcos/3  —  Psma 

/.  -=r  = 7=5 — : — 77-: — n =  coefficient  of  friction  =  /. 

R        —  Q  sin  /)  -f-  P  cos  a  J 

Let  the  resultant  of  F  and  R  make  an  angle  0  with  the 
normal  to  the  plane.  Then 


_F_          Q  cos  ft  —  P  sin  a  Q  _  sin  (a  +  0) 

~R~     -  Qsin  P  +  Pcosa'  P  ~  cos  (ft  -  $)' 

0  is  called  the  angle  of  friction.  It  has  also  been  called  the 
angle  of  repose,  since  a  body  will  remain  at  rest  on  an  inclined 
plane  so  long  as  its  inclination  does  not  exceed  the  angle  of 
friction. 


If  a  =  o  =  ft,     then     ^  =  tan  0  =  /. 

The  work  done  in  traversing  a  distance  x  =  Q  cos  ft .  x.     If 
Q  is  variable,  the  work  done  —    /     Q  cos  ft .  dx 

3.  Wedge. — The  wedge,  or  key,  is  often  employed  to  con- 
nect members  of  a  structure,  and  is  generally  driven  into  posi- 


WEDGE. 


303 


tion  by  the  blow  of  a  hammer.     It  is  also  employed  to  force 
out  moisture  from  materials  by  induc- 
ing a  pressure  thereon. 

The  figure  represents  a  wedge  de- 
scending vertically  under  a  continuous 
pressure  P,  thus  producing  a  lateral 
motion  in  the  horizontal  member  C, 
\yhich  must  therefore  exert  a  pressure 
Q  upon  the  vertical  face  AB. 

The  member  H  is  fixed,  and  it  is 
assumed  that  the  motion  of  the  machine 
is  uniform,  so  that  the  wedge  and  £7  are 
in  a  state  of  relative  equilibrium. 

Let  Rl ,  R^  be  the  reactions  at  the  faces  DE,  DF,  respec- 
tively, their  directions  making  an  angle  0,  equal  to  the  angle 
of  friction,  with  the  normals  to  the  corresponding  faces. 

Let  a  be  the  angle  between  DE  and  the  vertical,  a'  the 
angle  between  DF  and  the  vertical. 

Consider  the  wedge,  and  neglect  its  weight,  which  is  usually 
inappreciable  as  compared  with  P. 

Resolving  vertically, 


Rl  cos  (90°  —  a  +  0)  +  RS  cos  (90°  —  a' 

=  P=R1  sin  (a  +  0)  +  R,  sin  (af  -\-  0). 

Resolving  horizontally, 


or 


R,  sin  (90°  —  a  +  0)  —  R^  sin  (90°  —  a'  +  0)  =  o, 

Rl  cos  (or  +  0)  =  ^2  cos  (a'  +  0) (2) 


Consider  the  member  C,  and  neglect  its  weight. 
Resolving  horizontally, 

R,  cos  (a  +  0)  =  Q  =  R,  cos  (af  +  0).       . 


(3) 


Assuming  the  wedge  isosceles,  as  is  usually  the  case,  a  =  a', 
and  hence, 


by  eq.  (2),  Rl  =  Rtt  and  by  eq.  (i),  2R,  sin  (a  +  0)  =  P.       (4) 


3O4  THEORY  OF  STRUCTURES. 

Hence,  by  eqs.  (3)  and  (4), 

Q  _  cot  (OL  +  0)  __  external  resistance  overcome 

P~  2  effort  exerted  "  '5' 

(N.B. — This  ratio  of  resistance  to  effort  is  termed  the  mechan- 
ical advantage,  or  put  chase,  of  a  machine.) 

Suppose  the  motion  of  the  machine  reversed,  so  that  Q  be- 
comes the  effort  and  P  the  resistance. 

The  reactions  R, ,  R^  now  fall  below  the  normals,  and  the 
equations  of  relative  equilibrium  are  the  same  as  the  above, 
with  —  0  substituted  for  0. 


Thus,  —  =  £  cot  (a  —  0) (6) 

The  two  cases  may  be  included  in  the  expression 

-p  =*cot(«±0) (7) 


For  a  given  value  of  P,  Q  increases  with  a. 
If  there  were  no  friction,  0  would  be  zero,  and  eq.  (7)  would 
become 

Q       cot  a 


Thus,  the  effect  of  friction  may  be  allowed  for,  by  assuming 
the  wedge  frictionless,  but  with  an  angle  increased  by  20  in  the 
first  case,  and  diminished  by  20  in  the  second  case. 

Again,  when  P  is  the  effort  and  Q  the  resistance,  eq.  (5) 

shows  that  if  a  -\-  0  >  90°,  the  ratio  -y  is  negative,  which  is 
impossible,  while  if  a  -\-  0  =  90°,  -~   is  zero,  and  in  order  to 


WEDGE.  305 

overcome  Q,  however  small  it  might  be,  P  would  require  to  be 
infinitely  great.     Hence, 

a  +  0  must  be  <  90°, 

Q 
and  below  this  limit  -p  diminishes  as  0  increases. 

Similarly,  it  may  be  shown  from  eq.  (7)  that  when  Q  is  the 
effort  and  P  the  resistance, 

0  must  be  <  a, 

and  that  below  this  limit  5-  increases  with  0. 

Efficiency. — During  the  uniform  motion  of  the  machine,  let 
any  point  a  descend  vertically  to  the  point  b.  The  correspond- 
ing horizontal  displacement  is  evidently  2bc. 

The  motive  work  =  P .  ab  ; 
"    useful  work    =Q.2bc. 

Hence,  the  efficiency  =  ~ — 7-  =  -7,- .  2   tan  a 

•=  tan  a  cot  (a  -f-  0),  by  eq.  (5). 

This  is  a  maximum  for  a  given  value  of  0  when 


and    the   max.   efficiency  =  tan  ^45° j  cot  ^45°  +  -J 


For  the  reverse  motion,  the  efficiency 
P.ab 


3o6 


THEORY  OF  STRUCTUKES. 


0 

This  is  a  maximum  when  a  =.  45°  -{-  — .     Thus  the 


imax.  efficiency  =  cot  ^45°  -f-  -j  tan  145°  •  -—)  =  -• 


—  sin  0 

-{-  sin  0' 


4.  Screws. — A  screw  is  usually  designed  to  produce  a 
linear  motion  or  to  overcome  a  resistance  in  the  direction  of  its 
length.  It  is  set  in  motion  by  means  of  a  couple  acting  in  a 
plane  perpendicular  to  its  axis.  A  reaction  is  produced  be- 
tween the  screw  and  nut  which  must  necessarily  be  equivalent 
to  the  couple  and  resistance,  the  motion  being  steady. 

Take  the  case  of  a  square  *  -threaded  screw.  It  may  be 
assumed  that  the  reaction  is  concentrated  along  a  helical  line, 
whose  diameter,  d,  is  a  mean  between  the  external  and  internal 
diameters  of  the  thread,  and  that  its  distribution  along  this 
line  is  uniform.  It  will  also  be  supposed  that  the  axes  of  the 
couple  and  screw  are  coincident,  so  that  there  will  be  no  lateral 
pressure  on  the  nut. 

Let  M  be  the  driving  couple. 
"     Q   "     "     axial   resistance  to    be  over- 
come. 

"  r  "  "  reaction  at  any  point  a  of  the 
helical  line,  and  let  0  be 
angle  between  its  direction 
and  the  normal  at  a ;  0  is 
the  angle  of  friction. 

"      a   "     "     angle    between    the    tangent 
at  a  and  the  horizontal  ;  a 
is  called  the  pitch-angle. 
Since  the  reaction   between  the   screw   and   nut    must    be 
equivalent  to  M  and  Q,  then 

*  Square-threaded  screws  work  more  accurately  than  those  with  a  V-thread, 
but  the  efficiency  of  the  latter  has  been  shown  to  be  very  little  less  than  that  of 
the  former  (Poncelet).  On  the  other  hand,  the  V-thread  is  the  stronger,  much 
less  metal  being  removed  in  cutting  it  than  is  the  case  with  a  square  thread. 
Again,  with  a  V-thread  there  is  a  tendency  to  burst  the  nut,  which  does  not 
obtain  in  a  screw  with  a  square  thread. 


FIG.  240. 


SCRE  W  'S.  SO/ 

Q  —  algebraic  sum  of  vertical  components  of  the  reac- 
tions at  all  points  of  the  line  of  contact, 

=  2\r  cos  (a  +  0)]  =  cos  (a  +  <t>)S(r),  ....     (i) 

and  M  =  algebraic  sum  of  the  moments  with  respect  to  the 
axis  of  the  horizontal  components  of  the  reactions  at  all  points 
of  the  line  of  contact, 

.      .     .     (2) 

Let  the  couple  consist  of  two  equal  and  opposite  forces,  P, 
acting  at  the  ends  of  a  lever  of  length/,  so  that  M  =  Pp. 
Hence,  by  eqs.  (i)  and  (2), 


and  the  mechanical  advantage 


Q      ip 

If  0  =  o,    -p  =  —  cot  of,  and  the  effect  of  friction  may 

be  allowed  for,  by  assuming  the  screw  frictionless,  but  with  a 
pitch-angle  equal  to  a  -\-  0. 

Again,  let  the  figure  represent  one  complete  turn  of  the 
thread  developed  in  the  plane  of  the 
paper.   CD  is  the  corresponding  length 
of  the  thread  ;  DE  the  circumference 
it  d;   CE,  parallel  to  the  axis,  the  pitch   5" 
h  ;  and  CDE  the  pitch-angle  a.  F»G.  241. 

The  motive  work  in  one  revolution          =  M  .  2?r  =  Pp  .  2n. 

The  useful  work  done  in  one  revolution  =  Qh. 

Hence,  the  efficiency  =  ~^  =  22-  cot  (a  +  0)  j^ 

=     cot  (a  +  0)  =  tan  a  cot  (a  +  0)-   •   (4) 


3O8  THEORY   OF   STRUCTURES. 

0 

This  is  a  maximum  when  a  =  45°  —  -,  its  value  then  being 


In  practice,  however,  a  is  generally  much  smaller,  efficiency 
being  sacrificed  to  secure  a  large  mechanical  advantage,  which, 
according  to  eq.  (3),  increases  as  a  diminishes. 

If  a  -\-  0  =  90°,  —  =  o,  so  that  to  overcome  Q,  however 
small  it  may  be,  would  require  an  infinite  effort  P. 


Suppose  the  pitch-angle  sufficiently  coarse  to  allow  of  the 
screw  being  reversed.  Q  now  becomes  the  effort  and  P  the 
resistance.  The  direction  of  r  falls  on  the  other  side  of  the 
normal,  and  the  relation  between  P  and  Q  is  the  same  as 
above,  —  0  being  substituted  for  0. 

Thus, 


and  therefore  the  mechanical  advantage 


P 

If  a  =  0,  —  =  o,  and    to    overcome  P,  however  small   it 

may  be,  Q  would  require  to  be  infinite. 

.-.  a  >  0. 

If  a  <  0,  reversal  of  motion  is  impossible,  and  the  screw 
then  possesses  the  property,  so  important  in  practice,  of  serv- 
ing to  fasten  securely  together  different  structural  parts,  or  of 
locking  machines. 


ENDLESS  SCREWS. 


309 


Again,  it  may  be  necessary  to  take  into  account  the  friction 
between  the  nut  and  its  seat,  as  well  as  the  friction  at  the  end 
of  the  screw.  The  corresponding  moments  of  friction  with 
respect  to  the  axis  are  (Art.  8) 


f 

7 


3  4°  - 


and      /— ( 


/  being  the  coefficient  of  friction,  d, ,  d^  the  external  and  inter- 
nal diameters  of  the  seat,  and  d'  the  diameter  of  the  end  of 
the  screw. 

5.  Endless  Screws  (Fig.   242).— A  screw   is  often  made 
to  work  with  a   toothed   wheel,  as,  for  ex- 
ample, in  raising  sluice-gates,  when  the  screw 
is  also  made  sufficiently  fine  to  prevent,  by 
friction   alone,  the  gates  from   falling  back 
under  their  own  weight.     The  theory  is  very 
similar  to  the  preceding.    Let  the  screw  drive.    ( 
A  tooth  rises  on  the  thread,  and  the  wheel 
turns  against  a  tangential  resistance  Q,  which 
is  approximately  parallel  to  the  axis  of  the 


screw. 


FIG.  242. 


Let  Fig.  243  represent  one  complete  turn  of  the  thread 
developed  in  the  plane  of  the  paper,  a 
being  the  pitch-angle  as  before. 

Consider  a  tooth.  It  is  acted  upon 
by  Q  in  a  direction  parallel  to  the 
axis,  and  by  the  reaction  R  between 
the  thread  and  tooth,  making  an  angle 
0  (the  angle  of  friction)  with  the  normal 
to  the  thread  CD. 


FIG.  243. 


.-.  Q  =  R  cos  (a  +  0). 

Again,  the  horizontal  component  of  R,  viz.,  R  sin  (a  -|-  0), 
has  a  moment  R  sin  (a  -f-  0)  —  with  respect  to  the  axis  of  the 


3IO  THEORY  OF  STRUCTURES. 

screw,  and  this  must  be  equivalent  to  the  moment  of  the  driv- 
ing-couple, viz.,  Pp  (Art.  4). 


=  *    sin  (a 


Thus  the  relation  between  P  and  Q  is  the  same  as  in  the  pre 
ceding  article. 

Similarly  if  the  wheel  acts  as  the  driver, 


6.  Rolling  Friction.  —  The  friction  between  a  rolling  body 
and  the  surface  over  which  it  rolls  is  called  rolling  friction. 
Prof.  Osborne  Reynolds  has  given  the  true  explanation  of  the 
resistance  to  rolling  in  the  case  of  elastic  bodies.  The  roller 
produces  a  deformation  of  the  surfaces  in  contact,  so  that  the 
distance  rolled  over  is  greater  than  the  actual  distance  between 
the  terminal  points.  This  he  verified  by  experiment,  and  con- 
cluded that  the  resistance  to  rolling  was  due  to  the  sliding  of 
one  surface  over  the  other,  and  that  it  would  naturally  increase 
or  diminish  with  the  deformation.  In  proof  of  this  he  found, 
for  example,  that  the  resistance  to  an  iron  roller  on  india- 
rubber  is  ten  times  as  great  as  the  resistance  when  the  roller  is 
on  an  iron  surface.  Hence  the  harder  and  smoother  the  sur- 
faces, the  less  is  the  rolling  friction.  The  resistance  is  not 
sensibly  affected  by  the  use  of  lubricants,  as  the  advantage  of 
a  smaller  coefficient  of  friction  is  largely  counteracted  by  the 
increased  tendency  to  slip.  Other  experiments  are  yet  re- 
quired to  show  how  far  the  resistance  is  modified  by  the 
speed. 

Generally,  as  in  the  case  of  ordinary  roadways,  the  resist- 
ance is  chiefly  governed  by  the  amount  of  the  deformation  of 
the  surface  and  by  the  extent  to  which  its  material  is  crushed. 
Let  a  roller  of  weight  W  (Y\g.  244)  be  on  the  point  of  motion 
under  the  action  of  a  horizontal  pull  R. 


ROLLING  FRICTION.  31 1 

The  resultant  reaction  between  the  surfaces  in  contact 
must  pass  through  the  point  of  intersection  of  R  and  IV. 
Let  it  also  cut  the  surface  in  the  point  B. 

Let  d  be  the  horizontal  distance  between  B  and  W. 
11    p      "       vertical  "  "        B    "     R. 

Taking  moments  about  By 

Rp  =  Wd, 
or  R< 

R  =  the  resistance  =  W-. 

P 


Coulomb  and  Morin  inferred,  jw 

as  the  results  of  a  series  of  ex-  FIG.  244. 

periments,  that  d  is  independent  of  the  load  upon  the  roller  as 
well  as  of  its  diameter,*  but  is  dependent  upon  the  nature  of 
the  surfaces  in  contact. 

*  Dupuit's  experiments  led  him  to  the  conclusion  that  d  is  proportional  to 
the  square  root  of  the  diameter,  but  this  requires  further  verification. 
Let  n  be  the  coefficient  of  sliding  friction. 
The  resistance  of  the  roller  to  sliding  is  /j,  W,  and  "  rolling  "  will  be  insured 

d 
if  R  <  fj.  W,  i.e.,  if  -  <  tan  0,  which  is  generally  the  case  so  long  as  the  direc- 

/ 
tion  of  R  does  not  fall  below  the  centre  of  the  roller. 

Assume  that  R  is  applied  at  the  centre.  The  radius  r  may  be  substituted 
for/,  since  (/is  very  small,  and  hence 

R  =  W-. 
r 

An  equation  of  the  same  form  applies  to  a  wheel  rolling  on  a  hard  roadway 
over  obstacles  of  small  height,  and  also  when  rolling  on  soft  ground.  In  the 
latter  case,  the  resistance  is  proportional  to  the  product  of  the  weight  upon  the 
wheel  into  the  depth  of  the  rut,  and  the  depth  for  a  small  arc  is  inversely  pro- 
portional to  the  radius. 

Experiments  on  the  tractional  resistance  to  vehicles  on  ordinary  roads  are 
few  in  number  and  incomplete,  so  that  it  is  impossible  to  draw  therefrom  any 
general  conclusion. 

From  the  experiments  carried  out  by  Easton  and  Anderson,  it  would  appear 
that  the  value  of  d  in  inches  varies  from  1.6  to  2.6  for  wagons  on  soft  ground, 
and  that  the  resistance  is  not  sensibly  affected  by  the  use  of  springs.  Upon 
a  hard  road,  in  fair  condition,  the  resistance  was  found  to  be  irom  £  to  \  of 
that  on  the  soft  ground,  the  average  value  of  d  being  \  inch,  and  was  very 
sensibly  diminished  by  the  use  of  springs. 


312 


THEORY  OF  STRUCTURES. 


7.  Journal-friction. — Experiments  indicate  that  f  is  not 
the  same  for  curved  as  for  plane  surfaces,  and  in  the  ordinary 

cases  of  journals  turning  in  well- 
lubricated  bearings  the  value  of /is 
probably  governed  by  a  combina- 
tion of  the  laws  of  fluid  friction  and 
of  the  sliding  friction  of  solids. 

The  bearing  part  of  the  journal 
is  generally  truly  cylindrical  and  is 
terminated     by    shoulders     resting 
against   the    ends    of     the   step   in 
which  the  journal  turns. 
Consider  a  journal  in  a  semicircular  bearing  with  the  cap 
removed.     When  the  cap  is  screwed   on,  the  load  upon  the 
journal  will  be  increased  by  an  amount  approximately  equal 
to  the  tension  of  the  bolts.     Let  -Pbe  the  load. 

Assume  that  the  line  of  action  of  the  load  is  vertical  and 
that  it  intersects  the  axis  of  the  shaft.  This  load  is  balanced 
by  the  reaction  at  the  surface  of  contact,  but  much  uncertainty 
exists  as  to  the  manner  in  which  this  reaction  is  distributed. 
There  are  two  extremes,  the  one  corresponding  to  a  normal 
pressure  of  constant  intensity  at  every  point  of  contact,  the 
other  to  a  normal  pressure  of  an  intensity  varying  from  a 
maximum  at  the  lowest  point  A  to  a  minimum  at  the  edge  of 
the  bearing  B. 

Let  /  be  the  length  of  the  bearing,  and  consider  a  small 
element  AS  at  any  point  C,  the  radius  OC  (=  r)  making  an 
angle  6  with  the  vertical  OA. 

First.  Let  /  be  the  constant  normal  intensity  of  pressure. 


p  = 


e.  i)  — 


—  2pir. 


Frictional  resistance  = 


=  fpl^(AS}=fplnr—fP-  . 


The   frictional   resistance    probably  approximates    to    this 
limit  when  the  journal  is  new. 


JO  URN  A  L-  FRICTION.  3  1  3 

Second.  Let  /  =  pQ  cos  0, 

so  that  the  intensity  is  now  proportional  to  the  depth  CD  and 
varies  from  a  maximum  /0  at  A  to  nil  at  B.  This,  perhaps, 
represents  more  accurately  the  pressure  at  different  points 
when  the  journal  is  worn. 


—  2/0/r,    A  cosa  0  .  dO  =  /0/r- 

t/o  2 


and  the  frictional  resistance  =  2(fj>4Sl)  —  2fpjr  =  fP-. 
Hence,  the  frictional  resistance  lies  between  fP-    and  fP—  . 

2  7t 

It  may  be  represented  by  j*Pt  IJL  being  a  coefficient  of  friction 
to  be  determined  in  each  case  by  experiment. 

The  total  moment  of  frictional  resistance  must  necessarily 
be  equal  and  opposite  to  the  moment  M  of  the  couple  twisting 
the  shaft  ;  i.e., 

M  = 


Thus,  the  total  reaction  at  the  surface  of  contact  is  equiva- 
lent to  a  single  force  P  tangential  to  a  circle  of  radius  /*r  having 
its  centre  at  O  and  called  the  friction-circle. 

The  work  absorbed  by  axle-friction  per  revolution 


=  M.27t  •=.  2 
The  work  absorbed  by  axle-friction  per  minute 


N  being  the  number  of  revolutions  and   v  the  velocity  per 
minute. 


314  THEORY  OF  STRUCTURES. 

The  work  absorbed  by  frictional  resistance  produces  an 
equivalent  amount  of  heat,  which  should  be  dissipated  at  once 
in  order  to  prevent  the  journal  from  becoming  too  hot.  This 
may  be  done  by  giving  the  journal  sufficient  bearing  surface 
(an  area  equal  to  the  product  of  the  diameter  and  the  length 
of  the  bearing),  and  by  the  employment  of  a  suitable  unguent. 

Suppose  that  h  units  of  heat  per  square  inch  of  bearing 
surface  (Id)  are  dissipated  per  minute. 

Let  /  inches  be  the  length  and  d  inches  the  diameter  of 
the  journal. 

hdl  =  heat-units  dissipated  =  heat-units  equivalent  to 
frictional  resistance 


J  being  Joule's  equivalent,  or  778  ft.-lbs. 

12/ft        PN  \2jJl        Pv 

.*.  -  =  —r-    and     --  =  —  . 

fJLTt  I  ^  Id 

p 

Let  7-7  =p  =  pressure  per  square  inch  of  bearing  surface. 


Id 

\2jh 


=  a  constant. 


In  Morin's  experiments  af  varied  from  2  to  4  in.,  P  from 
330  Ibs.  to  2  tons,  and  v  did  not  exceed  30  ft.  per  minute;  so 
that/z/  was  <  5000,  and  the  coefficient  of  friction  for  the  given 
limits  was  found  to  be  the  same  as  for  sliding  friction. 

Much  greater  values  of  pv  occur  in  modern  practice. 

Rankine  gives  p(v  +  20)  =  44800  as  applicable  to  locomo- 
tives. 

Thurston  gives  pv  =  60000  as  applicable  to  marine  engines 
and  to  stationary  steam-engines. 

Frictional  wear  prevents  the  diminution  of  /below  a  certain 


JOURNAL   FRICTION.  31$ 

limit  at  which  the  pressure  per  unit  of  bearing  surf  ace  exceeds 
a  value/  given  by  the  formula. 


where 


- 


In  practice  k  =  %  for  slow-moving  journals  (e.g.,  joint-pins), 
and  varies  from  I-J  to  3  for  journals  in  continuous  motion.  The 
best  practice  makes  the  length  of  the  journal  equal  to  four 
diameters  (i.e.,  k  =  4)  for  mill-shafting. 

Again,  'if  the  journal  is  considered  a  beam  supported  at 
the  ends, 


q  being  the  maximum  permissible  stress  per  square  inch,  and 
C  a  coefficient  depending  upon  the  method  of  support  and 
upon  the  manner  of  the  loading. 

„     k 
/.  #  oc  —  . 

i 

For  a  given  value  of  P,  d  diminishes  as  q  increases.  Also, 
it  has  been  shown  that  the  work  absorbed  by  friction  is 
directly  proportional  to  d. 

Hence,  for  both  reasons,  d  should  be  a  minimum  and  the 
shaft  should  be  made  of  the  strongest  and  most  durable 
material.  In  practice  the  pressure  per  square  inch  of  bearing 
surface  may  be  taken  at  about  2  tons  per  square  inch  for  cast- 
iron,  3!  tons  per  square  inch  for  wrought-iron,  and  6J  tons  per 
square  inch  for  cast-steel. 

It  would  appear,  however,  from  the  recent  experiments  of 
Tower  and  others,  that  the  nature  of  the  material  might  become 
of  minor  importance,  while  that  of  a  suitable  lubricant  would  be 
of  paramount  importance.  They  show  that  the  friction  of 
properly  lubricated  journals  follows  the  laws  of  fluid  friction 
much  more  closely  than  those  of  solid  friction,  and  that  the 


3l6  THEORY   OF  STRUCTURES. 

lubrication  might  be  made  so  perfect  as  to  prevent  any  ab- 
solute contact  between  the  journal  and  its  bearing.  The 
journal  would  therefore  float  in  the  lubricant,  so  that  there 
would  be  no  metallic  friction.  The  loss  of  power  due  to  fric- 
tional  resistance,  as  well  as  the  consequent  wear  and  tear,  would 
be  very  considerably  diminished,  while  the  load  upon  the 
journal  might  be  increased  to  almost  any  extent. 

Tower's  experiments  also  indicate  that  the  friction  dimin- 
ishes as  the  temperature  rises,  a  result  which  had  already  been 
experimentally  determined  by  Him.  It  was  also  inferred  by 
Hirn  that,  if  the  temperature  were  kept  uniform,  the  friction 
would  be  approximately  proportional  to  Vv,  and  Thurston 
has  enunciated  the  law  that,  with  a  cool  bearing,  the  friction  is 
approximately  proportional  to  Vv  for  all  speeds  exceeding 
100  ft.  per  minute. 

With  a  speed  of  150  ft.  per  minute  and  with  pressures  vary- 
ing from  100  to  750  Ibs.  per  square  inch,  Thurston  found  ex- 
perimentally that /"varied  inversely  as  the  square  root  of  the 
intensity  of  the  pressure.  The  same  law,  but  without  any 
limitations  as  to  speed  or  pressure,  had  been  previously  stated 
by  Hirn. 

8.  Pivots. — Pivots  are  usually  cylindrical,  with  the  circular 
edge  of  the  base  removed  and  sometimes  with  the  whole  of 
the  base  rounded.  Conical  pivots  are  employed  in  special 
machines  in  which,  e.g.,  it  is  important  to  keep  the  axis  of  the 
shaft  in  an  invariable  position.  Spherical  pivots  are  often 
used  for  shafts  subject  to  sudden  shocks  or  to  a  lateral  move- 
ment. 

(a)  Cylindrical  Pivots. — If  the  shafts  are  to  run  slowly,  the 
intensity  of  pressure  (/)  on  the  step  should  not  be  so  great  as 
to  squeeze  out  the  lubricant.  Reuleaux  gives  the  following 
rules : 

The  maximum  value  of  /  in  Ibs.  per  square  inch  should  be 
700  for  wrought-iron  on  gun-metal,  470  for  cast-iron  on  gun- 
metal,  and  1400  for  wrought-iron  on  lignum-vitae. 

For  rapidly-moving  shafts, 

d=c 


PIVOTS. 


317 


n  being  the  number  of  revolutions  per  minute,  c  a  coefficient 
to  be  determined  by  experiment  (=.0045), 
and  Pthe  load  upon  the  pivot. 

Suppose  the  surface  of  the  step  to  be 
divided  into  rings,  and  let  one  of  these 
rings  be  bounded  by  the  radii  x,  x  -J-  dx. 

In  one  revolution  the  work  absorbed 
by  the  friction  of  this  ring 


=   ji    .  2nx .  dx .  2nx. 


Hence  the  total  work  absorbed  in  one  revolution 


where 


FIG.  246. 


1?  -  d* 


and  dl  ,  d^  are  the  external  and  internal  diameters  of  the  sur- 
face in  contact. 

If  the  whole  of  the  surface  is  in  contact,  d^  =  o,  and  the 
work  absorbed  =  \^,nPdr 

Again,  the  moment  of  friction  for  the  ring 


.dx.x  = 


and  the  total  moment 


12 


If  d^  —  o,  the  moment  =  —  dl  . 

Thus,  in  both  cases,  the  work  absorbed  by  friction  =  27t 
times  the  moment  of  friction. 


THEORY  OF  STRUCTURES. 

Let  D  be  the  mean  diameter  of  the  surface  in  contact 


Let  2y  be  the  width  of  the  surface  in  contact  =  d^  —  d^  . 
Then 

work  absorbed  =  fj.nP  \D  -\-  —  J. 

Sometimes  shafts  have  to  run  at  high  speeds  and  to  bear 
heavy  pressures,  as,  e.g.,  in  screw-propellers  and  turbines.  In 
order  that  there  may  be  as  little  vibration  as  possible,  p  must 
be  as  small  as  practicable,  and  this  is  to  some  extent  insured 
by  using  a  collar-journal. 

Let  N  be  the  number  of  collars,  and  let  d^  ,  d^  be  the  exter- 
nal and  internal  diameters  of  a  collar. 

Then  work  absorbed  by  friction  per  revolution  per  collar 

L  —  2n  X  moment  of  friction. 


o 
According  to  Reuleaux,  the  mean  diameter  of  a  collar 


n 

~- 


n  being  the  number  of  revolutions  per  minute. 

Also,  the  width  of  surface  in  contact  =  dl  —  d^  =  .48 
and  the  maximum  allowable  pressure  per  square  inch 


(b)  Wear.  —  The  wear  at  any  point  of  the  elementary  ring 
must  necessarily  be  proportional  to  the  friction  ///>,  and  also  to 
the  amount  of  rubbing  surface  which  passes  over  the  point  in 
a  unit  of  time,  i.e.,  the  velocity  Ax  ;  A  being  the  angular  ve- 
locity of  the  shaft. 


PIVOTS.  319 

Hence,  the  wear  at  any  point  is  proportional  to 

(c)  Conical  Pivots.  —  As  before,  suppose 
the  surface  of  the  step  to  be  divided  into  a 
number  of  elementary  rings.  Two  cases  will 
be  discussed  : 

First.  Assume  that  the  normal  intensity 
of  pressure  p  at  the  surface  of  contact  is 
constant. 

Let  x,x-\-dx  be  the  distances  of  D  and 
E,  respectively,  from  the  axis.  IGZ 

The  total  moment  of  friction 


sin  a 


3  sin  a 

x^  ,  x^  being  the  radii  of  the  top  and  bottom  sections  of  the 
step. 

Also,  P,  the  total  load  on  the  pivot, 

=  /    pDE  sin  a  .  2nx  =  2np  I    xdx 

U  JC<t  V  X<i 

=  «P(x?  -  *.•)• 

2  U.P     X*  _  X  8 

Hence  total  moment  of  friction  =  —  -  --  -  -  -. 

3  sin  a  x?  —  x* 

Second.  Assume  that  the  wear  is  of  such  a  nature  that  every 
point,  e.g.,  D,  descends  vertically  through  the  same  distance. 
Thus,  the  normal  wear  a  sin  a, 


or  up    x  a  sn  a, 

or  px  oc  sin  a. 

In  the  present  case  a  is  constant,  and  hence  px  =  a  con- 
stant. 


320  THEORY   OF  STRUCTURES. 

Thus,  total  moment  of  friction 


sm  a 

=«-*••>• 


r*1 

=     I 

t/.ra 

/»*, 

I     dx  — 

IX  *f 


Also,  P  =    I    pDE  sin  a  .  inx 

t/.ra 

=  2npx 


Hence  total  moment  of  friction  =  —  :  -  (x.  +  #,). 

2  sm  av  * 

Schieles  Pivots.  —  The  object  aimed  at  in  these  pivots 
is  to  give  the  step  such  a  form  that  the  wear 
and  the  pressure  are  the  same  at  all  points. 

Let  0  be  the  angle  made  by  the  tangent  at 
-V--/D     any  point  of  the  step  with  the  axis. 

Let  y  be  the  distance  of  the  point  from 
the  axis.     Then 

py  a  sin  0; 

and  hence  if/  is  constant, 
y  a  sin  6     or    y  cosec  0  =  a  const. 

is  the  equation  of  the  generating  line  of  the  step.  This  line  is 
known  as  the  tractrix  and  also  as  the  anti-friction  curve.  If 
the  tangent  at  D  intersects  the  axis  in  T, 

DT  =  y  cosec  6  =  a  const. 

The  curve  may  be  traced  by  passing  from  one  point  to  an- 
other and  keeping  the  tangent  DT  of  constant  length. 
The  above  equation  may  be  written 

ds 

y—r-  =  a  const.  =  a, 
'  dy 


BELTS  AND  ROPES. 


321 


or 


ds       a  dy  I         fdy  \» 

~d~x  =~~  ~y  ~dx  =  V  l  +  w  ' 


which  may  be  easily  integrated,  the  result  being  the  analytical 
equation  to  the  curve,  viz., 


-f-  Va*  —  y  +  a  const. 


Schiele  or  anti-friction  pivots  are  suitable  for  high  speeds,  but 
have  not  been  very  generally  adopted. 

9.  Belts  and  Ropes. — Let  the  figure  represent  a  pulley 
movable  about  a  journal  at  O,  and  let  a  belt  (or  rope),  acted 
upon  by  forces  7, ,  72  at  the  ends,  embrace  a  portion  ABC 
of  the  circumference  subtending  an  angle  a  at  the  centre. 

In  order  that  there  may  be  motion  in  the  direction  of  the 
arrow,  7,  must  exceed  72  by  an  amount  sufficient  to  overcome 
the  frictional  resistance  along  the  arc  of  contact  and  the  resist- 
ance to  bending  due  to  the  stiffness  of  the  belt. 

Consider  first  the  frictional  resist- 
ance, and  suppose  the  belt  to  be  on  the 
point  of  slipping. 

Any  small  element  BB'  (—  ds)  of 
the  belt  js  acted  upon  by  a  pull  T  tan- 
gential to  the  pulley  at  B,  a  pull  T—  dT 
tangential  to  the  pulley  at  B' ,  and  by  a 
reaction  equivalent  to  a  normal  torceRds 
at  the  middle  point  of  BB' ,  and  a  tan- 
gential force,  or  frictional  resistance, 

j&ds. 

Let  the  angle  COB  =  0,  and  the  an- 
gle BOB'  =  dB.      , 
Resolving  normally, 


FIG.  249. 


7/n 

(T+  T-dT)sm Rds  -  o. 


.322  THEORY   OF  STRUCTURES. 

Resolving  tangentially, 

2  ^    ' 

-jj.  being  the  coefficient  of  friction. 

Now  dti   being    very  small,    sin —  is   approximately    •— , 

dB 
cos  —  is  approximately  unity,  and   small  quantities   of   the 

second  order  may  be  disregarded. 

Hence,  eqs.  (i)  and  (2)  may  be  written 

TdO  —  Rds  —  o,       ......     (3) 


=  ° •     (4) 

dT 


,     or 
Integrating, 


C  being  a  constant  of  integration. 

When  6  =  0,     T  =  T9,     and  hence     log.T;  =  C. 


or  =  «^  ........     (6) 

7    2 

When  0  =  a,     T  =  Tl  ,    and  hence 


^  being   the  number  2.71828,  i.e.,  the  base  of   the  Naperian 
system  of  logarithms. 


BELTS  AND  ROPES.  323 

If  a  is  increased  by  /?,  the  new  ratio  of  tensions  will  be 
&&  times  the  old  ratio ;  so  that  if  OL  increases  in  arithmetical 
progression,  the  ratio  of  tensions  will  increase  in  geometrical 
progression.  This  rapid  increase  in  the  ratio  of  the  tensions, 
corresponding  to  a  comparatively  small  increase  in  the  arc  of 
contact,  is  utilized  in  «"  brakes" 
for  the  purpose  of  absorbing 
surplus  energy.  For  example  : 

A  flexible  brake  consisting 
of  an  iron  or  steel  strap,  or, 
again,  of  a  chain,  or  of  a  series 
of  iron  bars  faced  with  wood 
and  jointed  together,  embraces 
about  three-fourths  of  the  cir- 
cumference of  an  iron  or  wooden 
drum.  One  end  of  the  brake  FlG-  25°' 

is  secured  to  a  fixed  point  0  and  the  other  to  the  end  B  of  a 
lever  AOB  turning  about  a  fulcrum  at  O.  A  force  applied  at 
A  will  cause  the  brake  to  clasp  the  drum  and  so  produce  fric- 
tion which  will  gradually  bring  the  drum  to  rest. 

Let  &9  be  the  angular  velocity  of  the  drum  before  the  brake 
is  applied. 

Let  /  be  the  moment  of  inertia  of  the  drum  with  respect  to 
its  axis. 

The  kinetic  energy  of  the  drum  = . 

When  the  brake  is  applied,  the  motion  being  in  the  direc- 
tion of  the  arrow,  let  the  greater  and  less  tensions  at  its  ends 
be  Tlt  T^t  respectively. 

Let  n  be  the  number  of  revolutions  in  which  the  drum  is 
brought  to  rest.  Then 

i/fi?a  =  (7;-  T^ndn, (8) 

d  being  the  diameter  of  the  drum. 

Also,  if  Pis  the  force  applied  at  A,  and  if  /  and  q  are  the 
perpendicular  distances  of  O  from  the  directions  of  P  and  T9, 
respectively, 

Pp=T# (9) 


3  24  THEOR  Y  OF  S  7^R  UCTURES. 

Again, 

T*  =  TS*,      .........  (10) 

<x  being  the  angle  subtended  at  the  centre  by  the  arc  of  contact. 
Hence,  by  eqs.  (8),  (9),  (10),  f 

\  _  qltf 


-  i)7td  .....     '     ' 

If  the  motion  of  the  drum  were  in  the  opposite  direction,  q 
would  be  the  perpendicular  distance  of  O  from  the  direction  of 
T,  ,  and  then  Pp  =  T,q. 

Proceeding  as  before, 


_ 


and  therefore  the  number  of  turns  in  the  second  case,  before 
the  drum  comes  to  rest,  is  e^a  times  the  number  in  the  first, 
which  is  consequently  the  preferable  arrangement. 

The  coefficient  of  friction  JJL  varies  from  .12  for  greasy  shop 
belts  on  iron  pulleys  to  .5  for  new  belts  and  hempen  ropes  on 
wooden  drums.  In  ordinary  practice,  an  average  value  of  >u 
for  dry  belts  on  iron  pulleys  is  .28,  and  for  wire  ropes  .24;  if 
the  belts  are  wet,  >u  is  about  .38. 

Formulae  (6)  and  (7)  are  also  true  for  non-circular  pulleys. 

10.  Effective  Tension.  —  The  pull.available  for  the  trans- 
mission of  power  =  Tl  —  7!,  =  5.  Let  HP  be  the  horse- 
power transmitted,  v  the  speed  of  transmission  in  feet  per  sec- 
ond, a  the  sectional  area  of  the  rope  or  belt,  and  s  the  stress 
per  square  inch  in  the  advancing  portion  of  the  belt. 

Then,  if  Tl  and  Tt  are  in  pounds, 


=.,  and 

' 


, 

550      550' 

The  working  tensile  stress  per  square  inch  usually  adopted 
for  leather  belts  varies  from  285  Ibs.  (Morin)  to  35  5  Ibs.  (Claudel), 


EFFECT  OF  HIGH  SPEED.  325 

an  average  value  being  300  Ibs.  In  wire  ropes,  8500  Ibs.  per 
square  inch  may  be  considered  an  average  working  tension. 

Hempen  ropes  for  the  transmission  of  power  generally  vary 
from  4^  to  6J-  in.  in  circumference. 

II.  Effect  of  High  Speed.  —  When  the  speed  of  trans- 
mission is  great,  the  effect  of  centrifugal  force  must  be  taken 
into  account. 

wads  if 
The  centrifugal  force  or  the  element  ds  =  ----  ,  w  being 

the  specific  weight  of  the  belt  or  rope,  and  r  the  radius  of  the 
pulley. 

Eq.  (3)  above  now  becomes 

wads  v* 
Tdd  —  Rds  ----  =  o, 

g     r 

or 


and  hence,  by  eq.  (4), 


Integrating, 


-r        Wa 

T  —  — 


since  T  =  T^  when  6  =  o. 

Also,  T  =  T,  when  6  =  a,  and  therefore 


g 

or 


_        V«-  i). 

o 


326  THEORY  OF  STRUCTURES. 

The  work  transmitted  per  second 


=  (T,  -  7>  =    T,v  -      *>(<-  -  i), 


which    is    a    maximum    and    equal    to    |7'8(^a  —  i)    when 
,  and  the  two  tensions  are  then  in  the  ratio  of 


2^«  _(_  i  to  3. 

The  speed  for  which  no  work  is  transmitted,  i.e.,  the  limit- 
ing speed,  is  given  by 


wa  ,  i  /•*« 

\v  —  — v  =  o,      or      v  =  \f  — 


12.  Slip  of  Belts.  —  A  length  /  of  the  belt  (or  rope)  becomes 
/(  i  -j-^jon  the  advancing  side  and  l[\  +  -jj  on  the  slack  side, 


T  T 

where  pl  =  —  -  and  /„  =  —-  ,  E  being  the  coefficient  of  elasticity. 

Thus,  the  advancing  pulley  draws  on  a  greater  length  than  is 
given  off  to  the  driven  pulley,  and  its  speed  must  therefore 
exceed  that  of  the  latter  by  an  amount  given  by  the  equation 


ifi]  _ 
reduction  of  speed,  or  slip  _  ~  El         \     r  E'  _  pl  — • 

speed  of  driving  pulley  /        p 


The  slip  or  creep  of  the  belt  measures  the  loss  of  work. 
In  ordinary  practice  the  loss  with  leather  belting  does  not  ex- 
ceed 2  per  cent,  while  with  wire  ropes  it  is  so  small  that  it  may 
be  disregarded. 


PA  0  N  Y '  S  D  YNA  MO  ME  TER. 


327 


13.  Prony's  Dynamometer. — This  dynamometer  is  one  of 
the  commonest  forms  of  friction-brake.  The  motor  whose 
power  is  to  be  measured  turns  a  wheel  E  which  revolves  be- 
tween the  wood  block  B  and  a  band  of  wood  blocks  A.  To 


/INP 


FIG.  251. 


the  lower  block  is  attached   a  lever  of  radius  /  carrying  a 

weight  P  at  the  free  end.     By  means  of  the  screws  C,  D  the 

blocks  may  be  tightened  around  the  circumference  until  the 

unknown  moment  of  frictional  resistance  FR  is  equal  to  the 
known  moment  Pp. 

The  weight  P,  which  rests  upon  the  ground  when  the 
screws  are  slack,  is  now  just  balanced. 

The  work  absorbed  by  friction  per  minute  =  2nRFn  =  2nPpnv 

n  being  the  number  of  revolutions  per  minute. 

14.  Stiffness  of  Belts  and  Ropes. — The  belt  on  reaching 
the  pulley  is  bent  to  the  curvature  of  the  periphery,  and  is 
straightened  again  when  it  leaves  the  pulley.  Thus,  an  amount 
of  work,  increasing  with  the  stiffness  of  the  belt,  must  be  ex- 
pended to  overcome  the  resistance  to  bending.  As  the  result 
of  experiment,  this  resistance  has  been  expressed  in  the  form 

T-B,  T  being  the  tension  of  the  belt,  a  its  sectional  area,  R  the 

radius  of  the  pulley,  and  b  a  coefficient  to  be  determined. 

According  to  Redtenbacher,  b  =  2.36  in.  for  hempen  ropes. 
"  "  "  b  =  1.67  "     "         "  " 

"  "  Reuleaux,  b  =  3.4     "     "    leather  belts. 


328 


THEORY  OF  STRUCTURES. 


Let  the  figure  represent  a  sheave  in  a  pulley-block  turning 
in  the  direction  of  the  arrow  about  a 
journal  of  radius  r. 

Let  Tt  be  the  effort,  T9  the  re- 
sistance. 

The  resistance  due  to  the  stiff- 
ness of  the  belt  may  be  allowed  for 

*T* 

qTs  by  adding  -j-£  to  the  force  7",  .     The 
ftR  OK 

FIG.  252.  frictional  resistance  at  the  journal- 

surface  is  P  sin  0  orfP,  P  being  the  resultant  of  Tlt  71,  . 

The  motion  being  steady,  taking  moments  about  the  centre, 


or 


If  T;  and  T9  are  parallel,  P  =  T>  +  T>,  and  the  last  equa 
tion  becomes 


Let  the  pulley  turn  through  a  small  angle  6. 
The  counter-efficiency  of  the  sheave 


motive  work       Tfl  _  T^ 
useful  work  z=         = 


2fr 


,  a 
~^~ 


R  -  fr~~  b  R  -  fr' 

In  the  case  of  an  endless  belt  connecting  a  pair  of  pulleys 
of  radius  R19  R^  the  resistance  due  to  stiffness  may  be  taken 

equal  to  ^-\jf  "f"j>"/*  ^Demg  tne  mean  tension  [=  ~~L~~^ 

The  resistance  due  to  journal-friction  =  frP\-£  +^~J- 
The  useful  resistance  =  T,  —  T^  =  S. 


WHEEL   AND  AXLE. 

Hence,  the  counter-efficiency 


329 


In  wire  ropes  the  stress  due  to  bending  may  be  calculated 
as  follows: 

Let  x  be  the  radius  of  a  wire.  The  radius  of  its  axis  is 
sensibly  the  same  as  the  radius  R  of  the  pulley. 

The  outer  layers  of  the  wire  will  be  stretched,  and  the  inner 
shortened,  while  the  axis  will  remain  unchanged  in  length. 

Hence, 

x       change  of  length  of  outer  or  inner  strands      unit  stress 
R  ~  length  of  axis  E 

x 
and  the  unit  stress  due  to  bending  =  £~ . 

15.  Wheel  and  Axle. — Let  the  figure  represent  a  wheel 
of  radius  /  turning  on  an  axle  of  radius  r,  under  the  action  of 
the  two  tangential  forces  P  and  Q,  in- 
clined to  each  other  at  an  angle  6. 

The  resultant  R  of  P  and  Q  must 
equilibrate  the  resultant  reaction  be- 
tween the  wheel  and  axle  at  the  sur- 
face of  contact. 

Let  the  directions  of  P  and  Q 
meet  in  T. 

If  there  were  no  friction,  the  re- 
sultant reaction  and  the  resultant  R 
would  necessarily  pass  through  O 
and  T. 

Taking  friction  into  account,  the 
direction  of  R  will  be  inclined  to  TO. 
Let  its  direction  intersect  the  circumference  of  the  axle  in  the 
point  A.  The  angle  between  TA  and  the  normal  AO  at  A, 
the  motion  being  steady,  is  equal  to  the  angle  of  friction  ;  call 
it  0. 


FIG.  253. 


330  THEORY  OF  STRUCTURES. 

Taking  moments  about  O, 

Pp  —  Qp  —  Rr  sin  0  =  o 


Also, 


.  .  (I) 

R*  =  P*  +  Q  +  2PQ  cos  e (2) 


Let/=  sin  0  = 


=,  fjL  being  the  coefficient  of  friction. 


Eq.  (i)  may  now  be  written 

Pp-  Qp-fRr  =  0.    .     , 
'    If  P  and  Q  are  parallel  in  direction, 

8  =  0    and     R  =  P  -4-  Q. 
Let  the  figure  represent  a  wheel  and  axle. 


(3) 


254. 


Let  P  be  the  effort  and  Q  the  weight  lifted,  the  directions 
of  Pand  Q  being  parallel. 

Let  IV  be  the  weight  of  the  "  wheel  and  axle." 

Let  R1  and  R^  be  the  vertical  reactions  at  the  bearings. 

Let/  be  the  radius  of  the  wheel. 

Let  q      "  "         "        axle. 

Let  r      "  "         "        bearings. 

Take  moments  about  the  axis.     Then 

Pp  —  Qq  —  RS  sl'n  0  ~  RS  sin  0  =  0.    .    .    (4) 


TOOTHED   GEARING.  331 

But 


......    (5) 

Hence, 


or 

P(p-fr)  =  Q(q+fr)+fWr.    ........     (6) 

Efficiency.  —  In  turning  through  an  angle  0, 
motive  work  =  PpQ, 
useful   work  ==  QqQ, 


effirier  - 

"e  -~     =     ' 


and  the  ratio  -p  is  given  by  eq.  (6). 

16.  Toothed  Gearing.  —  In  toothed  gearing  the  friction  is 
partly  rolling  and  partly  sliding,  but  the  former  will  be  disre- 
garded, as  it  is  small  as  compared  with  the  latter. 


Let  the  pitch-circles  of  a  pair  of  teeth  in  contact  at  the 
point  B  touch  at  the  point  A  ;  and  consider  the  action  before 
reaching  the  line  of  centres  0a0a,  i.e.,  along  the  arc  of 
approach. 


332  THEORY  OF  STRUCTURES. 

The  line  AB  is  normal  to  the  surfaces  in  contact  at  the 
point  B. 

Let  R  be  the  resultant  reaction  at  B.  Its  direction,  the 
motion  being  steady,  makes  an  angle  0,  equal  to  the  angle  of 
friction,  with  AB. 

Let  B  be  the  angle  between  O,Ot  and  AB. 

Let  the  motive  force  and  force  of  resistance  be  respectively 
equivalent  to  a  force  P  tangential  to  the  pitch-circle  Ol  ,  arid 
to  a  force  Q  tangential  to  the  pitch-circle  O,  . 

Let  rl  ,  ra  be  the  radii  of  the  two  wheels. 

The  work  absorbed  by  friction  in  turning  through  the  small 
arc  ds 

.    (i) 


Consider  the  wheel  Ov  ,  and  take  moments  about  the  centre, 
Pr1  =  ^|r1sin(6/-0)  +  ^sm0f,      ...     (2) 

where  AB  =  x. 

Similarly,  from  the  wheel  <93 

0ra  =  tf  K  sin  (0  -  0)  -  *  sin  0}.      ...     (3) 

Hence, 

jtr 
n      sin  (0  -  0)  —  -  sin  4> 

=  -  --     -  ......     (4) 


sin  (0  —  0)  -f-  -  sin  0 
f  i 

and  therefore 

'i        i  \ 

(r  +  rl*  sin  0 


'i 

+     *  sn  0 

(5) 


sin  (0  —  0)  --  sin  0 


Hence,  the  work  absorbed  by  friction  in  the  arc  ds 

=  Q- 

sin  (6  —  0) sin  0 

7*0 


TOOTHED   GEARING.  333 

In  precisely  the  same  manner  it  can  be  shown  that,  after 
leaving  the  line  of  centres,  i.e.,  in  the  arc  of  recess, 

ft       sin  (0  -|-  0)  —  —  sin  0 


p  -  x         » 

sin  (0+0)  +  -  sin  0 


and  the  work  absorbed  by  friction  in  the  arc  ds 


=  Q—  —  —  --  .    ....     (8) 

sin  (0+0)  --  sin  0 

^2 

The  ratio  -p  and  the  loss  of  work  given  by  eqs.  (4)  and  (6) 

are  respectively  greater  than  the  ratio  ^~  and  the  loss  of  work 

given  by  eqs.  (7)  and  (8),  and  therefore  it  is  advisable  to  make 
the  arc  of  approach  as  small  as  possible. 

Again,  by  eq.  (4),  motion  will  be  impossible  if 

sin  (6  —  0)  +  -  sin  0  =  o  ; 
**i 

i.e.,  if  cot  0  =  cot  0  — 


» 
r1  sin  0 

and  this  can  only  be  true  if  the  direction  of  R  passes  through  (93  . 

Simple  approximate  expressions  for  the  lost  work  and 
efficiency  may  be  obtained  as  follows: 

6  differs  very  little  from  90°,  and  x  is  small  as  compared 
with  r2  and  differs  little  from  the  corresponding  arc  s  meas- 
ured from  A. 

Hence  the  work  absorbed  by  friction  in  the  arc  ds 


=  Q  tan  0      -        *&  = 


334  .THEORY  OF  STRUCTURES. 

and  the  work  lost  in  arc  of  approach  s1 


The  useful  work  done  in  the  same  interval  =  Qsl  . 
The  counter-efficiency  (reciprocal  of  efficiency) 


Similarly  for  the  arc  of  recess  s^  , 


the  lost  work  =  g/J-  +  -—  ,      -     •     (i  i) 

*7*j        T'j/  2 

and  the  counter-efficiency  =  i  -}-/*(—  -J--  )—.      .     (12) 

\r,      /y  2 

If  ^  =  ,ya  =  pitch  =  /  =  -  -1  =  -  ?,  n^  ,  «2  being  the  num- 

^1  ?^3 

her  of  teeth  in  the  driver  and  the  follower,  respectively,  the  ex- 
pressions for  the  lost  work  given  by  eqs.  (9)  and  (11)  are  iden- 
tical, and  those  for  the  counter-efficiency  given  by  eqs.  (10) 
and  (12)  are  also  identical. 

Thus,  the  whole  work  lost  during  the  action  of  a  pair  of 
teeth 


.......   (13) 

and  the  counter-efficiency 


<•» 


This  last  equation  shows  that  the  efficiency  increases  with 
the  number  of  teeth. 


EFFICIENCY  OF  MECHANISMS. 


335 


If  the  follower  is  an  annular  wheel,  -  —  -  must  be  substi- 


tuted  for  -+-  in  the  above  equations.  Thus,  with  an  an- 
nular wheel  the  counter-efficiency  is  diminished  and  the 
efficiency,  therefore,  increased. 

It  has  been  assumed  that  R  and  Q  are  constant,  as  their 
variation  from  a  constant  value  is  probably  small.  It  has  also 
been  assumed  that  only  one  pair  of  teeth  are  in  contact.  The 
theory,  however,  holds  good  when  more  than  one  pair  are  in 
contact,  an  effort  and  resistance,  corresponding  to  P  and  Q, 
being  supposed  to  act  for  each  pair. 

17.  Bevel-wheels. — Let  I  A,  IB  represent   the    develop- 
ments  of    the   axes   of    the   pitch- 
circles  77, ,  773  of   a  pair  of  bevel- 
wheels   when    the    pitch-cones   are 

spread    out  fiat,  Ol ,    Oz   being  the 
corresponding  centres. 

The  preceding  formulae  will  ap- 
ply to  bevel-wheels,  the  radii  being 
6^7,  <927,  and  the  pitch  being  meas-  ,!,'''' 
ured  on  the  circumferences  I  A,  IB. 

18.  Efficiency  of  Mechanisms. 
— Generally  speaking,  the   ratio  of 
the  effort  P  to  the  resistance  Q  in  a. 
mechanism  may  be  expressed  as  a 
function  of   the  coefficient  of   fric- 
tion yu.     Thus, 


t 


If,  now,  the  mechanism  is  moved  so  that  the  points  of 
application  of  P  and  Q  traverse  small  distances  Ax,  Ay  in  the 
directions  of  the  forces, 


.1.       ^  . 

the  efficiency  =  j~-  = 
PAx 


I     Ay 


336 


THEORY   OF  STRUCTURES. 


Ay 

But  the  ratio  -r-  depends  only  upon  the  geometrical  rela- 
tions between  the  different  parts  of  the  mechanism,  and  will 
therefore  remain  the  same  if  it  is  assumed  that  /*  is  zero.  In 
such  a  case  the  efficiency  would  be  perfect,  or  the  motive  work 
(PAx)  would  be  equal  to  the  useful  work  (QAy\  and  therefore 


I  = 


I     Ay 


Hence,  the  efficiency 


TABLE  OF  COEFFICIENTS   OF  AXLE-FRICTION. 


>, 

Q 

Greasy  and 
Wet. 

Ordinary 
Lubrication. 

Continuous 
Lubrication. 

Pure  Carriage- 
grease. 

Lard  and 
Plumbago. 

O 

Bell-metal  on  bell-metal  

Brass  on  brass    . 

Brass  on  cast-iron 

Cast-iron  on  bell-metal  

161 

o6<; 

.16 

Cast-iron  on  brass                    , 

Cast-iron  on  cast-iron  

•  14 

Cast-iron  on  lignum-vitae. 

xSe 

Lignum-vitae  on  cast-iron  
Lignum-vitae  on  lignum-vitae  
Wrought-iron  on  bell-metal  
Wrought-iron  on  cast-iron  

.251 

.189 

.n6 
•075 

•J7 
.07 
•  054 

.09 

.11 

•15 

Wrought-iron  on  lignum-vitae.   .  . 

.187 

•125 

EXAMPLES.  337 


EXAMPLES. 


i.  In  a  pair  of  four-sheaved  blocks,  it  is  found  that  it  requires  a  force 
P  to  raise  a  weight  $P',  and  a  force  $P  to  raise  a  weight  i  $P'.  Show 
that  the  general  relation  between  the  force  P  and  the  weight  W  to  be 
raised  is  given  by 


Find  the  efficiency  when  raising  the  weights  5/"  and 

2.  Find  the  mechanical  advantage  when  an  inch  bolt  is  screwed  up 
by  a  i5-in.  spanner,  the  effective  diameter  of  the  nut  being  if  in.,  the 
diameter  at  the  base  of  the  thread  .84  in.,  and  .15  being  the  coefficient  of 
friction. 

3.  A  belt,  embracing  one-half  the  circumference  of  a  pulley,  transmits 
10  H.  P.  ;  the  pulley  makes  30  revolutions  per  minute  and  is  7  ft.  in 
diameter.     Neglecting  slip,  find  TV  and  Ti  ;  /a  being  .125. 

4.  A  £-in.  rope  passes  over  a  6-in.  pulley,  the  diameter  of  the  axis  being; 
%  in.  ;  the  load  upon  the  axis  ==  2  x  the  rope  tension.     Find  the  efficiency 
of  the  pulley,  the  coefficient  of  axle-friction  being  .08  and  the  coefficient 
for  stiffness  .47. 

Hence  also  deduce  the  efficiency  of  a  pair  of  three-sheaved  blocks. 

5.  If  the  pulleys  are  50  ft.  c.  to  c.  and  if  the  tight  is  three  times  the 
slack  tension,  find  the  length  of  the  belt,  the  coefficient  of  friction  being 
£  and  the  diameter  of  one  of  the  pulleys  12  in. 

6.  Show  that  the  work  transmitted  by  a  belt  passing  over  a  pulley 

/~T 
will  be  a  maximum  when  it  travels  at  the  rate  of  A/  _£_2  ft.  per  sec.,  T* 

yn 
being  the  slack  tension  and  m  the  mass  of  a  unit  of  length  of  the  belt. 

The  tight  tension  on  a  2o-in.  belt,  embracing  one-half  the  circum- 
ference of  the  pulley,  is  1200  Ibs.  Find  the  maximum  work  the  belt  will 
transmit,  the  thickness  of  the  belt  being  .2  in.  and  its  weight  .0325  Ib. 
per  cubic  inch.  (Coefficient  of  friction  =  .28.) 

7.  In  an  endless  belt  passing  over  two  pulleys,  the  least  tension  is 
150  Ibs.,  the  coefficient  of  friction  .28,  and  the  angle  subtended  by  the 
arc  of  contact  148°.    Find  the  greatest  tension.     The  diameter  of  the 
larger  wheel  is  78  in.,  of  the  smaller   10  in.,  of  the  bearings  3  in.    Find 
the  efficiency.     A  tightening-pulley  is  made  to  press  on  the  slack  side 
of  the  belt.     Assuming  that  the  working  tension  is  to  the  coefficient  of 
elasticity  in  the  ratio  of  i  to  80,  find  the  increment  of  the  arc  of  contact 


THEORY  OF  STRUCTURES. 

on  the   belt-pulley,  the  tension  of  the  slack  side,  and  the  force  of  the 
tightening-pulley. 

8.  A  belt  weighing  £•  Ib.  per  lineal  foot,  connects  two  42-in.  pulleys,  one 
making  240  revolutions  per  minute.    Find  the  limiting  tension  for  which 
•work  will  be  transmitted.     Also  find  the  tight  and  slack  tensions  and 
the  efficiency  when  the  belt  transmits  5  horse-power.     Diameter  of  axle 
—  2  in.;  coefficient  of  friction  =  .28. 

9.  A  circular  saw  makes  1000  revolutions  per  minute  and  is  driven 
"by  a  belt  3  in.  wide  and  \  in.  thick,  its°weight  per  cubic  inch  being  .0325 
Ib.     The  belt  passes  over  a  lo-in.  pulley,  embracing  one-half  the   cir- 
cumference, and  transmits  6  H.  P.     Find  the  light  and  slack  tensions, 
the  coefficient  of  friction  being  .28. 

10.  A  flexible  band,  embracing  three-fourths  of  the  circumference  of 
a  brake-pulley  keyed  on  a  revolving  shaft,  has  one  extremity  attached  to 
the  end  A  of  the  lever  AOB,  and  the  other  to  \\\t  fixed  point  O  (between 
A  and  J3)  about  which  the  lever  oscillates.     The  pressure  between  the 
band  and  pulley  is  effected  by  a  force  applied  at  right  angles  to  the  lever 
at  the  end  B.     Show  that  the  time  in  which  the  axle  is  brought  to  rest 
is  about  i\  times  as  great  when  revolving  in  one  direction  as  in  the 
opposite  (/  =  .2). 

11.  In  a  Prony-brake  test  of  a  Westinghouse  engine,  the  blocks  were 
fixed  to  a  24-in.  fly-wheel  with  a  6-in.  face,  and  the  balance-reading  was 
48  Ibs.;  the  distance  from  centre  of  shaft  to  centre  of  balance,  measured 
horizontally,  was  30  in.,  and  the  number  of  revolutions  per  minute  was 
624.    Find  the  H.  P.  Ans.  14.3. 

12.  An  engine  makes  150  revolutions  per  minute.    If  the  diameter  of 
the  brake-pulley  is  45  in.  and  the  pull  on  the  brake  is  50  Ibs.,  find  the 
B.  H.  P.  Ans.  2.67. 

13.  A  small  water-motor  is  tested  by  a  tail  dynamometer.     The  pul- 
ley is  18  in.  in  diameter;  the  weight  is  60  Ibs.;   the    spring  registers   a 
pull  of  50  Ibs.;   the  number  of  revolutions  per  minute  =  500.    Find  the 
B.  H.  P.  Ans.  f 

14.  The  power  of  an  engine  making  n  revolutions  per  minute  is 
tested  by  a  Prony  brake  having  its  arm  of  length  r  connected  with  a 
spring-balance  which  registers  a  force  P.     The  arm  is  vertical  and  the 
weight  W  of  the  brake  is  supported  by  a  stiff  spring  fixed  vertically 
below  the  centre  of  the  wheel.     What  error  in  B.  H.  P.  would  be  intro- 
duced by  placing  the  spring  x  ft.  away  from  the  central  position  ? 

Ans.   Bl^*  ,  B  being  the  B.  H.  P. 

15.  Find  work  absorbed  by  friction  per  revolution  by  a  pivot  3  in. 
long  and   carrying  6  tons,  its  upper  face  being  6  in.  in  diameter,  coeffi- 
cient of  friction  .04,  and  2  a  being  90°. 


EXAMPLES.  339 

1 6.  The  diameter  of  a  solid  cylindrical  cast-steel  pivot  is  2%  in.    Find 
the  diameter  of  an  equally  efficient  conical  pivot. 

17.  The  pressure  upon  a  4-in.  journal  making  50  revolutions  per  min- 
ute is  6  tons,  the  coefficient  of  friction  being  .05.     Find  the  number  of 
units  of  heat  generated  per  second;  Joule's  mechanical  equivalent  of 
heat  being  778  ft.-lbs. 

1 8.  A  water-wheel  of  20  ft.  diameter  and  weighing  20,000  Ibs.  makes 
10  revolutions  per  minute;  the  gudgeons  are  6  in.  in  diameter  and  the 
coefficient  of  friction  is  .1.     Find  the  loss  of  mechanical  effect  due  to 
friction.     If  the  motive  power  is  suddenly  cut  off,  how  many  revolutions 
will  the  wheel  make  before  coming  to  rest  ?  Ans.  £f  H.  P. ;  10.9. 

19.  A  fly-wheel  weighing  8000  Ibs.  and  having  a  radius  of  gyration  of 
10  ft.  is  disconnected  from   the  engine  at  the  moment  it  is  making  27 
revolutions  per  minute  ;  it  stops  after  making  17  revolutions.     Find  the 
coefficient  of  friction,  the  axle  being  12  in.  in  diameter.         Ans.  .2325. 

20.  A  railway  truck  weighing  12  tons  is  carried  on  wheels  3  ft.  in 
diameter ;  the  journals  are  4  in.  in  diameter,  the  coefficient  of  friction 
•jjV    Find  the  resistance  of  the  truck  so  far  as  it  arises  from  the  friction 
of  the  journals.  Ans.  37^  Ibs. 

21.  A  tramcar  wheel  is  30  in.  in  diameter,  the  axle  2$-  in.;  the  coeffi- 
cient of  axle-friction  .08,  of  rolling  friction  .09.    Find  the  resistance  pet 
ton.  Ans.  28.37  Ibs. 

22.  A  bearing  16  in.  in  diameter  is  acted  upon  by  a  horizontal  force 
of  50  tons  and  a  vertical  force  of  10  tons ;  the  coefficient  of  friction  is  -fa. 
Find  the  H.  P.  absorbed  by  friction  per  revolution.       Ans.  .906  H.  P. 

23.  A  steel  pivot  3  in.  in  diameter  and  under  a  pressure  of  5  tons 
makes  60  revolutions  per  minute  in  a  cast-iron  step  well  lubricated  with 
oil.    How  much  work  is  absorbed  by  friction,  the  coefficient  of  friction 
being  .08  ? 

24.  A  pair  of  spur-wheels  are  4  in.  and  2  in.  in  diameter  ;  the  flanks 
of  the  teeth  are  radial ;  the  larger  wheel  has  16  teeth  ;  the  arc  of  ap- 
proach =  arc  of  recess  =  |  of  the   pitch.     Show  how  to  form  the  teeth, 
and  find  their  efficiency.    (Coefficient  of  friction  =  .n.) 

25.  Find  the  work  lost  by  the  friction  of  a  pair  of  teeth,  the  number 
of  teeth  in  the  wheels  being  32  and  16,  and  the  diameter  of  the  larger 
wheel,  which  transmits  3  horse-power  at  50  revolutions  per  minute,  3  ft. 

26.  The  driver  of  a  pair  of  wheels  has  120  teeth,  and  each  wheel  has 
an  addendum   equal   to  .28  times  the  pitch  ;    the  arcs  of  approach  and 
recess  are  each  equal  to  the  pitch  ;  the  tooth-flanks  are  radial.    (Coeffi- 
cient of  friction  =  .106.)     Find  the  efficiency. 


CHAPTER  VI. 


ON  THE  TRANSVERSE  STRENGTH  OF  BEAMS. 


FIG.  257. 


I.  To  determine  the  Elastic  Moment.  —  Let  the  plane  of 
--  |h  the  paper  be  a  plane  of  symmetry 
_  |c  with  respect  to  the  beam  PQRS. 

If  the  beam  is  subjected  to  the 
action  of  external  forces  in  this 
plane,  PQRS  is  bent  and  as- 
sumes a  curved  form  P'Q'R'S'. 
FIG.  258.  The  upper  layer  of  fibres,  Q'R',  is 

extended,  the  lower  layer,  P'S',  is  compressed,  while  of  the 
layers  within  the  beam,  those  nearer  P'S'  are  compressed  and 
those  nearer  Q'R'  are  extended.  Hence,  there  must  be  a  layer 
M'N'  between  P'S'  and  Q'R'  which  is  neither  compressed  nor 
extended.  It  is  called  the  neutral  surface  (or  cylinder),  and 
its  axis  is  perpendicular  to  the  plane  of  flexure.  In  the  present 
treatise  it  is  proposed  to  deal  with  flexure  in  one  plane  only, 
and,  in  general,  it  will  be  found  more  convenient  to  refer  to 
M'N'  as  the  neutral  line  (or  axis),  a  term  only  used  in  refer- 
ence to  a  transverse  section. 

If  a  force  act  upon  the  beam  in  the  direction  of  its  length, 
the  lower  layer  P'S',  instead  of  being  compressed,  may  be 
stretched.  In  such  a  case  there  is  no  neutral  surface  within 
the  beam,  but  theoretically  it  still  exists  some- 
where without  the  beam. 

Let  ABCD  be  an  indefinitely  small  rect- 
angular element  of  the  unstrained  beam,  and 
let  its  length  be  s.  Let  A'B'C'D',  Fig.  260, 
be  the  element  after  deformation  by  the  external  forces. 

340 


FIG.  259. 


THE  ELASTIC  MOMENT.  34* 

P'Q',  the  neutral  line,  being  neither  com-    , 
pressed  nor  extended,  is  unchanged  in  length 


and  equal  to  PQ  —  s. 

Let  the  normals  at  Pr  and  Q'  to  the  neutral 
line  meet  in  the  point  O  ;  O  is  the  centre  of 
curvature  of  P'Q'* 

Also,  as  the  flexure  of  the  element  is  very 
small,  the  normal  planes  through  OP'  and  OQ' 
may  be  assumed  to  be  perpendicular  to  all  the 
layers  which  traverse  the  corresponding  sec- 
tions of  the  beam,  so  that  they  must  coincide 
with  the  planes  A  '  D'  and  B  '  C'  ',  respectively. 

The  assumptions  made  in  the  above  are  : 

(a)  That   the    beam    is    symmetrical   with 
respect  to  a  certain  plane. 

(b)  That  the  material  of  the  beam  is  homo- 
geneous. FIG.  260. 

(c)  That  sections  which  are  plane  before  bending  remain 
plane  after  bending. 

(d)  That  the  ratio  of  longitudinal  stress  to  the  correspond- 
ing strain  is  the  ordinary  (i.e.,  Young's)  modulus  of  elasticity 
notwithstanding    the    lateral   connection    of   the    elementary 
layers. 

(e)  That    these   elementary   layers   expand    and    contract 
freely  under  tensile  and  compressive  forces. 

Consider  an  elementary  layer  p'q',  of  length  s',  sectional 
area  al  ,  and  distant  yl  from  the  neutral  surface. 
Let  OP'  =  R  =  OQ'. 
From  the  similar  figures  OP'Q  and  Op'q', 

Op'        p'q'  R+y,      s'  yv      s'-s 


Also,  if  tl  is  the  stress  along  the  layer  p'q'  , 


342  THEORY  OF  STRUCTURES. 

E  being  the  coefficient  of  elasticity  of  the  material    of^  the 
beam. 

So,  if  /2 ,  #2 ,  72 ,  /3 ,  a3 ,  jj/3 ,  .  .  .  are  respectively  the  stress, 
sectional  area,  and  distance  from  the  neutral  surface,  of  the 
several  layers  of  the  element, 


The  total  stress  along  the  beam  is  the  algebraic  sum  of  all 
these  elementary  stresses, 


Again,  the  moment  of  /,  about  P'  =  t^  =      ^7^  ; 


and  so  on. 

Thus,  the  Elastic  Moment  for  the  section  A ' D'  =  the  alge- 
braic sum  of  the  moments  of  all  the  elementary  stresses  in  the 
different  layers  about  P'9 


THE  ELASTIC  MOMENT.  343 

Now,  2  (ay*)  is  the  moment  of  inertia  of  the  section  of  the 
beam  through  A  'D  ',  with  respect  to  a  straight  line  passing 
through  the  neutral  line  and  perpendicular  to  the  plane  of 
flexure,  i.e.,  the  plane  of  the  paper.  It  is  usually  denoted  by 
/  or  AJ?,  A  being  the  sectional  area,  and  k  the  radius  of 
gyration.  Thus, 

E         E 

the  elastic  moment  =  -7,-  /  =  -^Ak*. 
K.  K. 

But  the  elastic  moment  is  equal  and  opposite  to  the  bending 
moment  (M)  due  to  the  external  forces,  at  the  same  section. 
Hence 


Note.—\\.  is  necessary  in  the  above  to  use  the  term  alge- 
braic, as  the  elementary  stresses  change  in  character,  and 
therefore  in  sign,  on  passing  from  one  side  of  the  neutral  sur- 
face to  the  other. 

Cor.  I.  Bearing  in  mind  assumption 
(e),  the  figure  represents  on  an  exaggerated 
scale  the  transverse  section  of  the  beam 
at  A'D  ',  the  upper  and  lower  breadths 
of  the  beam,  A'  A"  and  H  D"  ,  being  re- 
spectively contracted  and  stretched,  and 
being  also  arcs  of  circles  having  a  common 
centre  at  O'. 

Let  R'  be  the  radius  of  the  arc  P'  P"  , 
whose  length  remains  unchanged. 

Let   mE  be  the  lateral  coefficient  of 
FlG-  2fii-  elasticity,  m  being  a  numerical  coefficient. 

As  before,  for  any  layer  at  a  distance  y  from  P'P", 


/.  R'  =  mR. 


344  THEORY  OF  STRUCTURES. 

Thus,  within  the  limits  of  elasticity,  the  curvature  of  the  breadth 

is  —  that  of  the  length,  and  does  not  sensibly  affect  the  re- 

sistance of  the  beam  to  bending.  The  influence,  however, 
upon  the  bending  may  become  sensible  if  the  breadth  is  very 
large  as  compared  with  the  depth,  as,  e.g.,  in  the  case  of  iron 
or  steel  plates. 

Cor.  2.  If  the  resolved  part  of  the  external  forces  in  the 
direction  of  the  length  of  the  beam  is  nil, 

E 
the  total   longitudinal  stress  =  -^(ay)  =  o,  or  ^(ay)  =  o, 


showing  that  Pf  must  be  the  centre  of  gravity  of  the  section 
through  A'  D'  .  Hence,  when  the  external  forces  produce  no 
longitudinal  stress  in  the  beam,  the  neutral  line  is  the  locus  of 
the  centres  of  gravity  of  all  the  sections  perpendicular  to  the 
length  of  the  beam. 

Cor.  3.  If  /,  a,  y  be,  respectively,  the  stress,  sectional  area, 
and  distance  of  a  fibre  from  the  neutral  line,  then 


F  E         t 

--ay  =  /,   or   -=,-y  =  -  =  intensity  of  stress  =fy,  suppose, 

J\.  rL          a 


EXAMPLE  i.  A  timber  beam,  6  in.  square  and  20  ft.  long, 
rests  upon  two  supports,  and  is  uniformly  loaded  with  a  weight 
of  1000  Ibs.  per  lineal  foot.  Determine  the  stress  at  the  centre 
at  a  point  distant  2  in.  from  the  neutral  line. 

Also  find  the  central  curvature,  E  being  1,200,000  Ibs. 


/T    /^ 

/=-—  =  108,    M  =  1000  X  io—  1000  X  5  =  5000    ft.-lbs. 
=  60,000  inch -Ibs.,     and    y  =  2  in. 


INTERNAL    STRESSES.  345 

Hence  from  the  above  equations, 


1200000  fy 

— —  x  108  =  60000  =  — 


Thus  R  =  2160  in.  =  180  ft.,     and    fy—  nii-fr  Ibs.  per  sq.  in. 

Ex.  2.  A  standpipe  section,  33  ft.  in  length  and  weighing 
5720  Ibs.,  is  placed  upon  two  supports  in  the  same  horizontal 
plane,  30  ft.  apart.  The  internal  diameter  of  the  pipe  is  30 
in.,  and  its  thickness  j-  inch.  Determine  the  additional 
uniformly  distributed  load  which  the  pipe  can  carry  between 
the  bearings,  so  that  the  stress  in  the  metal  may  nowhere  ex- 
ceed 2  tons  per  square  inch. 

Let  J^be  the  required  load  in  pounds. 

3° 

The  weight  of  the  pipe  between  the  bearings  =  —  .  5720 

=  5200  Ibs. 

Thus,  the  total  distributed  weight  between  the  bearings 
=  (^+5200  Ibs.) 

Now  M=f 


and  the  stress  in  the  metal  is  necessarily  greatest  at  the  central 
section. 

W+  5200 
M,  at  the  centre,  =  — .30.  12  inch-lbs. ; 


fe  =  2  X  2240  Ibs.,     and     -  =  nr*t  =  —  .  1 5' ..  -. 


W-\-  52OO  22         ,     I 

— ^ .  30.  12  =   2.2240.--.  15    •-  =  72000  X  22, 

8  .72 


and  hence  W  —  30,000  Ibs. 


34-6  THEORY  OF  STRUCTURES. 

Cor.  4.  The  beam  is  strained  to  the  limit  of  safety  when 
either  of  the  extreme  layers  A'B',  D ' C  is  strained  to  the  limit 

of  elasticity.     In  such  a  case,  the  least  of  the  values  of  —  for 

y 

the  extreme  layers  A'B',  D' C  is  the  greatest  consistent  with 
the  strength  of  the  beam  ;  and  if  fc  and  c  are  the  corresponding 
intensity  of  stress,  and  distance  from  the  neutral  axis, 


EXAMPLE. — Compare  the  strengths  of  two  similarly  loaded 
beams  of  the  same  material,  of  equal  lengths  and  equal  sectional 
areas,  the  one  being  round  and  the  other  square. 

Let  r  be  the  radius  of  the  round  beam ;  fr9  the  intensity  of 
the  skin  stress. 

Let  a  be  a  side  of  the  square  beam ;  fa9  the  intensity  of 
the  skin  stress.  Then 


—  <22 ;    /,  for  round  bar,  = ,  and  for  square  bar  =  — . 


Also,  since  the  beams   are  similarly  loaded,  the   bending 
moments  at  corresponding  points  are  equal. 


r    4  #12 

2 

so  that 


Thus,  under  the  same  load,  the  round  beam  is  strained  to 
a  greater  extent  than  the  square  beam,  and  the  latter  is  the 
stronger  in  the  ratio  of  \^SS  to 


BREAKING    WEIGHTS.  347 

Cor.  5.  The  neutral  surface  is  neither  stretched  nor  com- 
pressed, so  that  it  is  not  subjected  to  any  longitudinal  stress. 
But  it  by  no  means  follows  that  this  surface  is  wholly  free  from 
stress,  and  it  will  be  subsequently  seen  that  the  effect  of  a 
shearing  force,  when  it  exists,  is  to  stretch  and  compress  the 
different  particles  in  diagonal  directions  making  angles  of  45° 
with  the  surface. 

bd*  d 

Cor.  6.  For  a  rectangular  beam//  =  — ,     and     c  =  — . 

,.*=// =/**=/«.. 

c          d   12       6 


If  the  beam  is  fixed  at  one  end  and  loaded  at  the  other 
with  a  weight  W,  the  maximum  bending  moment  =  Wl. 

If  the  beam  is  fixed  at  one  end  and  loaded  uniformly  with 
a  weight  wl  =  W,  the  maximum  bending  moment 

wP       Wl 


If  the  beam  rests  upon  two  supports  and  carries  a  weight  W 

Wl 
at  the  centre,  the  maximum  bending  moment  —  — , 

If  the  beam  rests  upon  two  supports  and  carries  a  uniformly 
distributed  load  of  wl  =  W,  the  maximum  bending  moment 
Wl 


Hence,  in  the  first  case,  IV  =  £  —=-\ 


"     "    second-     W=2 

"     «    third     «     W=^ 

"    "    fourth  "     W=^- 


THEORY   OF  STRUCTURES. 


In  general,  -  W  =  g-  q-j-  ; 


t/  being  some  coefficient  depending  upon  the  manner  of  the 
loading. 

Now,  if  the  laws  of  elasticity  held  true  up  to  the  point  of 
rupture,  these  equations  w^ould  give  the  breaking  weights  (W7), 
corresponding  to  different  ultimate  unit  stresses  (/),  but  the 
values  thus  derived  differ  widely  from  the  results  of  experi- 
ment. It  is  usual  to  determine  the  breaking  weight  (W)  of  a 

rectangular  beam  from  the  formula  W  =  C -T-,  where  C  is  a 

constant  which  depends  both  upon  the  manner  of  the  loading 
and  the  nature  of  the  material,  and  is  called  the  coefficient  of 
rupture. 

The  modulus  of  rupture  is  the  value  of/  in  the  ordinary 

bending-moment  formula  (M  =  — /)   when   the   load  on  the 

beam  is  its  breaking  load. 

The  preceding  equations,  however,  may  be  evidently  em- 
ployed to  determine  the  breaking  weights  in  the  several  cases 

by  making  J-^-q  =  C.     In  this  case /"is  no  longer  the  real  stress, 

but  may  be  called  the  coefficient  of  bending  strength. 

The  values  of  C  for  iron,  steel,  and  timber  beams,  supported 
at  the  two  ends  and  loaded  in  the  centre,  are  given. in  the 
Tables  at  the  end  of  Chapter  III. 

The  corresponding  value  of  f  is  obtained  from  the  equation 


or 


EXAMPLE.  —  Determine   the  central  breaking  weight  of   a 


EQUALIZATION  OF  STRESS.  349 

red-pine  beam,  10  in.  deep,  6  in.  wide,  and  resting  upon  two 
supports  20  ft.  apart. 

The  value  of  C  for  red  pine  is  about  5700.     Hence, 


the   breaking   weight  =  W  —  5700  — ll?_  =  14,250  Ibs. 

2O  X   12 


2.  Equalization  of  Stress. — The  stress  at  any  point  of  a 
beam  under  a  transverse  load  is  proportional  to  its  distance 
from  the  neutral  plane  so  long  as  the  elastic  limit  is  not  ex- 
ceeded. At  this  limit  materials  which  have  no  ductility  give 
way.  In  materials  possessing  ductility,  the  stress  may  go  on 
increasing  for  some  distance  beyond  the  elastic  limit  without 
producing  rupture,  but  the  stress  is  no  longer  proportional  to 
the  distance  from  the  neutral  plane,  its  variation  being  much 
slower.  This  is  due  to  the  fact  that  the  portion  in  compres- 
sion acquires  increased  rigidity  and  so  exerts  a  continually 
increasing  resistance  (Chap.  Ill)  almost  if  not  quite  up  to 
the  point  of  rupture,  while  in  the  stretched  portion  a  flow  of 
metal  occurs  and  an  approximately  constant  resistance  to  the 
stress  is  developed.  Thus,  there  will  be  a  more  or  less  perfect 
equalization  of  stress  throughout  the  section,  accompanied  by 
an  increase  of  the  elastic  limit  and  of  the  apparent  strength, 
the  increase  depending  both  upon  the  form  of  section  and  the 
ductility. 

For  example,  if  the  tensile  elastic  limit  is  the  same  as  the 
compressive.  the  shaded  portion  of  Fig.  262  gives  a  graphical 


__ 

^'-J== 


FIG.  262.  FIG.  263.  FIG.  264. 

representation  of  the  total  stress  in  a  beam  of  rectangular 
section  when  the  straining  is  within  the  elastic  limit.  Beyond 
this  limit,  it  may  be  represented  as  in  Fig.  263,  and  will  be 


350 


THEORY  OF  STRUCTURES. 


intermediate  between  Fig.  262  and  the  shaded  rectangle  of 
Fig.  264  which  corresponds  to  a  state  of  perfect  equaliza- 
tion. 

3.  Surface  Loading.* — It  may  be  well  to  draw  attention  to 
another  important  assumption  upon  which  is  based  the  mathe- 
matical treatment  of  the  problem  of  Beam  Flexure. 

It  has  been  assumed  that  the  external  forces  acting  on  a 
beam  can  be  so  applied  that  they  may  be  considered  as  dis- 
tributed uniformly  over  the  whole  section.  Thus  when  a  beam 
encastre"  is  loaded  at  the  free  end,  Fig.  265,  the  load  P  is  as- 


FIG.  265. 

sumed  to  be  uniformly  distributed  over  the  section  ab,  i.e., 
each  element  in  the  section  is  supposed  to  experience  the  same 
amount  of  strain  due  to  the  load,  and  the  reaction  of  the  wall 
is  also  supposed  to  be  uniformly  distributed  over  each  element 
in  the  section  cd. 

It  is  clear  that  such  suppositions  must  be  far  from  the 
truth. 

In  practice,  the  load  P  must  be  hung  by  some  means  from 
the  beam,  say  by  a  stirrup  passing  over  the  top.  The  whole 
load  is  then  concentrated  at  the  line  of  contact  of  the  stirrup 
with  the  beam,  and  it  is  obviously  untrue  to  say  that  every 


*  This  article  was  kindly  written    by  Professor   Carus-Wilson  and  is  an 
abstract  of  a  Paper  presented  by  him  to  the  Physical  Society. 


SURFACE  LOADING.  35 1 

element  in  the  section  ab  is  equally  strained.  But  more 
than  this.  It  has  been  assumed  that,  taking  the  effect  of 
the  load  as  distributed  uniformly  over  the  section  ab,  and 
a  certain  deflection  thereby  produced,  the  effect  of  P  on 
each  element  of  the  section  ab  may  be  disregarded  in  com- 
parison with  the  strains  involved  in  the  deflection  which  P 
produces. 

It  will  probably  be  difficult  at  first  to  grasp  the  fact  that 
certain  measurable  effects  have  been  actually  neglected,  but 
that  this  is  so  may  be  seen  by  supposing  the  beam  in  question 
to  be  a  pine  beam,  and  the  stirrup  of  iron.  Experience  proves 
that  with  a  very  moderate  load  the  beam  will  be  indented  at  a. 

But  the  theory  shows  that  the  longitudinal  tension  at  a  is 
zero  and  increases  to  a  maximum  at  d. 

Thus,  so  far  from  the  squeezing  effect  of  the  load  being 
distributed  uniformly  over  the  section  ab,  it  is  concentrated  at 
a,  and  hence  it  is  impossible  to  neglect  it. 

Engineers  have  always  recognized  the  existence  of  this 
"  surface-loading"  effect  in  practice,  and  where  possible,  have 
provided  a  good  "  bearing"  in  order  to  avoid  such  local 
strains ;  but  this  cannot  always  be  done — as,  for  instance, 
in  the  case  of  rollers  under  bridge  ends.  The  theory  of  flex- 
ure is  therefore  manifestly  incomplete  if  it  cannot  take  into 
account  the  actual  manner  in  which  the  loads  are  and  must  be 
applied. 


FIG.  266. 


It  can  be  shown  that  the  effect  of  placing  a  pressure  of  p 
tons  per  inch  run,  say  in  the  form  of  a  loaded  roller,  on  a  beam 
resting  upon  a  flat  surface,  as  in  Fig.  266,  to  prevent  it  from 


352 


THEORY   OF  STRUCTURES. 


bending,  is  to  compress  every  element  say  along  ab  with  an 
intensity  given  approximately  by  the  equation 


where  f  is  the  pressure  at  a  distance  x  from  a,  the  point  of  con- 
tact, and  h  =  ab.  This  is  the  equation  to  a  curve  be  which  is 
approximately  an  hyperbola. 

When  a  beam  is  bent  by  the  application  of  external  forces,  a 
very  close  approximation  to  the  true  condition  may  be  obtained 
by  superposing  this  surface-loading  effect  on  that  found  for 
bending. 

Take  the  case  of  a  beam  supported  at  the  ends  and  loaded 
at  the  centre,  and  let  it  be  required  to  find  the  condition  along 
ab,  Fig.  267. 


FIG.  267. 

The  effect  of  the  bending  is  to  produce  compression  above 
and  tension  below  the  point  c,  and  these  effects  may  be  repre- 
sented by  a  right  line  de  passing  through  c. 

The  surface-loading  effect  may  be  represented  by  an  hyper- 
bola giving  the  compression  at  any  point  along  ab  due  to  the 
load.  The  hyperbola  and  straight  line  will  intersect  in  two 


SURFACE   LOADING.  353 

points  h  and/,  which  shows  that  at  two  points  k ',/'  along  ab  the 
vertical  squeeze  produced  by  the  load  is  of  equal  intensity  to 
the  horizontal  squeeze  produced  by  the  bending ;  hence  an 
element  at  each  of  these  points  is  subject  to  cubical  compres- 
sion only.  From  a  to/'  the  beam  is  squeezed  vertically,  from 
/'  to  h'  it  is  squeezed  horizontally,  and  from^'  to  b  it  is  stretched 
horizontally.  The  intensities  are  given  at  every  point  by  the 
difference  between  the  ordinates  of  the  line  of  bending  de  and 
the  curve  of  loading.  It  will  appear  that  one  effect  of  surface- 
loading  is  to  make  the  neutral  axis  rise  up  under  the  load  and 
pass  through  the  point  hf,  for  there  is  neither  compression  nor 
tension  at  that  point. 

This  can  be  verified  by  examining  the  condition  of  a  bent 
glass  beam  by  polarized  light.  The  neutral  axis  is  pushed  up 
under  the  load  and  there  is  a  black  ring  passing  through  the  point 
/'.  If  the  span  is  diminished  and  the  load  kept  constant,  it  is 
clear  that  ae  will  become  less,  while  the  curve  of  loading  remains 
the  same,  until  the  line  dee  ceases  to  cut  the  curve ;  every 
element  along  ab  will  then  be  subjected  to  horizontal  stretch, 
and  the  stretch  is  greatest  at  a  ;  the  result  obtained  by  neglect- 
ing the  surface  loading  is  that  only  elements  from  c  to  b  are 
stretched,  the  greatest  stretch  being  at  b.  The  position  of  the 
"  neutral  points  "  is  given  by  the  equation 


T6~m 


where  y  is  the  distance  from  the  top  edge,  h  equals  the  depth 
ab,  m  —  -^-y 4,  and  a  =  one-half  of  the  span. 

For  all  elements  in  ab  to  be  stretched,  the  ratio  of  span  to 
depth,  viz.,  — -=— ,  must  be  equal  to  or  less  than  4.25.     In  other 

words,  for  any  beam,  and  any  load,  if  the  span  is  less  than  4^ 
times  the  depth,  every  element  in  the  normal  under  the  load 
is  stretched  horizontally. 


354 


THEORY   OF  STRUCTURES. 


4.  Beam  acted  upon  by  a  Bending  Moment  in  a  Plane 
which  is  not  a  Principal  Plane. 

Let*XOX,  YOYbe  the  principal  axes  of  the  plane  section 
^of  the  beam. 


\y 


FIG.  268. 

Let  the  axis  MOM  of   the  bending  moment  M  make  an 
angle  ot  with  OX. 

M  may  be  resolved  into  two  components,  viz., 

M  cos  a  —  X    and     M  sin  a  •=.  Y. 

These  components  may  be  dealt  with  separately  and  the 
results  superposed. 

Thus,  the  total  stress,  /,  at  any  point  (xy  y) 

—  stress  due  to  X  +  stress  due  to  Y  =  ~-  +  —  =/, 

1*  -fy 

I* ,  Iy  being  the  moments  of  inertia  with  respect  to  the  axes 
XOX,  YOY,  respectively. 

If  the  point  (xy)  is  on  the  neutral  axis,  then 


or 


being  the  angle  between  the  neutral  axis  and  XOX. 


SPRINGS. 


355 


Also  see  Art.  6,  Chap.  VIII.  In  this  article  6  is  the  angle 
between  the  neutral  axis  and  the  axis  of  the  couple,  i.e., 
6  =  ft  -  a. 

5.  Springs. — (a)  Flat  Springs. — If  two  forces,  each  equal 
to  P  but  acting  in  opposite  directions  in  the  same  straight  line, 
are  applied  to  the  ends  of  a  straight  uniform  strip  of  flat  steel 
spring,  the  spring  will  assume  one  of  the  forms  shown  below, 
known  as  the  elastic  curve.  This  curve  is  also  the  form  of  the 
linear  arch  best  suited  to  withstand  a  fluid  pressure,  Chap. 
XIII. 


FIG.  274. 


FIG.  275. 


Consider  a  point  B  of  the  spring  distant  y  from  the  line  of 
action  of  P.     Then 

Py  —  bending  moment  at  B  =  -=- , 


THEORY   OF  STRUCTURES. 

R  being  the  radius  of  curvature  at  B,  and  /  the  moment   of 
inertia  of  the  section. 

If  E  and  /are  both  constant, 

Ry  =  a  constant 

is  the  equation  to  the  elastic  curve. 

(b)  Spiral  Springs  (as,  e.g.,  in  a  watch).  —  Let  the  figure  rep- 
resent a  spiral  spring  fixed  at  C  and  to   an   arbor  at  A,  and 
subjected  at  every  point  of  its  length 
to  a  bending  action  only. 

Consider    the    equilibrium     of    any 
portion  AB  of  the  spring. 

The  forces  at  A  are  equivalent  to  a 
couple  of  moment  M,  and  to  a  force  P 
acting  in  some  direction  AD. 
This  couple  and  force  must   balance  the   elastic  moment 
at  B. 

.'.  M-\-  Py  =  El  X  change  of  curvature  at  B, 
y  being  the  distance  of  B  from  the  line  of  action  of  P,  or 


R0  being  the  radius  of  curvature  at  B  before  winding,  and  R 
that  after  winding. 

Let  ds  be  an  elementary  length  of  the  spring  at  B. 

Then,  for  the  whole  spring, 


2(M  +  Py]ds  =  £12       -     -    -  £12(40  - 


or    M2ds  -f-  P^yds  —El  X  total  change  of  curvature  between 
A  and  C  ; 

.-.  Ms  -\-  Psy  =  EI(B  -  00), 


SP  KINGS.  357 

s  being  the  length  of  the  spring,  y  the  distance  of  its  C.  of  G. 
from  AD,  6  the  angle  through  which  the  spring  is  wound  up, 
and  60  the  "  unwinding"  due  to  the  fixture  at  C.  With  a  large 
number  of  coils  the  distance  between  the  C.  of  G.  and  A  may 
be  assumed  to  be  nil  and  then  y  =  o. 

Also,  if  the  spring  is  so  secured  that  there  is  no  change  of 
direction  relatively  to  the  barrel, 

00  =  o,     and     Ms  =  Eld. 

Let  the  winding-up  be  effected  by  a  couple  of  moment 
Qq  =  M,  Q  being  a  tangential  force  at  the  circumference  of  a 
circle  of  radius  q. 

The  distance  through  which  Q  moves  (or  deflection  of  Q) 

=  g#  =       s,     since     M  =    f, 


/being  the  skin  stress,  and  c  the  distance  of  the  neutral  axis 
of  the  spring  from  the  skin. 

Thus,  if  b  is  the  width  of  a  spring  of  circular  or  rectangular 

b 
section,  c  =  — ,  and  hence 

the  deflection  =  -j~s. 
The  work  done  =  -Q  x  deflection.=  -~q6  = 

2  2    q  2 

/'  si     y 


2    E^       '2      EC*        '    2E      £a    ! 

k*  being  the  square  of  the  radius  of  gyration,  A  the  sectional 
area  of  the  spring,  and  Fits  volume. 

In  case  of  spring  of  rectangular  section  -r  =  -  . 

«     «      «        «        «    circuiar  «         4  =  -• 

*         4 


358 


THEORY  OF  STRUCTURES. 


Again,  the  spiral  spring  in  Fig.  277  is  wholly  subjected  to 
a  bending  action  by  means  of  a  twisting  couple  of  moment 
M  =  Qq  in  a  plane  perpendicular  to  the  axis  of  the  spring. 
Any  torsion  in  the  spring  itself  is  now  due  to  the  coils  not 
being  perfectly  flat. 


FIG.  277. 

Let  R0  =  radius  of  a  coil  before  the  couple  is  applied. 
"    R   =      "        "  "    "    after       "        "         u        " 


6  being  the  angle  of  twist  ;  or 


El     ~  El    ~  R~~  ^0~v 

N  being  the  number  of  coils  before  the  couple  is  applied,  and 
^y      "        «  «         «      «     after       "        "         "       u 

The  distance  through  which  g  acts,  i.e.,  the  "  deflection," 


and  the  work  done 


fV  V 


_ 

8   E 


for  spring  of  rectangular  section, 
"         u        "    circular 


6.  Beams  of  Uniform  Strength. — A  beam  having  the 
same  maximum  unit  stress  (/")  at  every  section  is  said  to  be  a 
beam  of  uniform  strength. 


BEAMS   OF    UNIFORM   STRENGTH. 


359 


At  any  section  of  a  beam  AB  (—  /)  denote  the  bending 
moment  by  M,  the  depth  of  the  beam  by  y,  and  its  breadth 
by  b.  Then 


£  being  the  distance  of  the  skin  from  the  neutral  axis,  and  A 
the  area  of  the  section. 

Evidently  c  and  k  are  each  proportional  to  y,  and  A  to  by. 


or 


.-.  fbf  cc  J/, 
nfbf  =  Mt 


n  being  a  coefficient  whose  value  depends  upon  the  form  of 
section. 

Four  cases  will  be  considered. 

CASE  a.  Assume  that  the  breadth  b  is  constant,  and  let 

nfb  =  j.     Then 


or 


y  =  ± 


Thus  AB  may  be  either  the  lower  edge  of  the  beam,  the 
ordinates  of  the  upper  edge  being  the  different  values  of  y,  or 
it  may  be  a  line  of  symmetry  with  respect  to  the  profile,  in 

which  case  the  ordinates  are  the  different  values  of  ±  -. 

EXAMPLE  I.  A  cantilever  AB  loaded  at  the  free  end  with  a 
weight  W^. 

At  a  distance  x  from  At 


y  =  pM  —  p  w,x. 

Theoretically,  therefore,  the 
beam,  in  elevation,  is  the  area 
ACD,  the  curve  CAD  being  a 


B 


FIG.  278. 


THEORY   OF  STRUCTURES. 

parabola   with    its   vertex    at    A    and    having    a   parameter 

The  max.  depth  =  2CB  —  CD  =  Vp  WJ. 

The  form  of  this  beam  is  very  similar  to  that  adopted  for 
cranks  and  for  the  cast-iron  beams  of  engines.  In  the  latter, 
the  material  is  usually  concentrated  in  the  flanges,  a  rib  being 
reserved  along  the  neutral  axis  for  purposes  of  connection. 

Again,  geometrical  conditions  of  transmission  require  the 
teeth  of  wheels  to  be  of  approximately  uniform  strength. 

A  cantilever  of  approximately  uniform  strength  may  be  ob- 
tained by  taking  the  tangents  C£,  DF  as  the  upper  and  lower 
edges  of  the  beam  instead  of  the  curves  CA,  DA.  The  depth 
of  the  beam  at  A  is  then  EF  —  \CD  =  j-  VpW~L  Although, 
theoretically,  the  depth  at  A  is  nil,  practically  the  beam  must 
have  sufficient  sectional  area  at  A  to  bear  the  shear  due  to  Wl , 
and  the  depth  £  VpWJ  will  be  found  ample  for  this  purpose. 

Note. — The  dotted  lines  show  the  beams  of  uniform 
strength,  when  the  lower  edge  is  the  horizontal  line  AB. 

Ex.  2.   A  cantilever   AB   carrying  a  uniformly  distributed 

load  W,. 

At  a  distance  x  from  A, 


or 


FIG.  279. 

The  beam,  in  elevation,  is  there- 

fore the  area  A  CD,  AC,  AD  being  two  straight  lines,  and  the 
maximum  depth  being 


The  sectional  area  at  A  is  nil,  as  both  the  bending  moment 
and  shear  at  that  point  are  zero. 


BEAMS  OF   UNIFORM   STRENGTH. 


361 


Note. — The  dotted  lines  show  the  cantilever  of  uniform 
strength  when  AB  is  the  lower  edge. 

Ex.  3.  A  cantilever  AB  carrying  a  weight  Wl  at  the  free 
end  A  and  also  a  uniformly  distributed  load  W^, 


FIG.  280. 

At  the  distance  x  from  A, 


This  equation  may  be  written  in  the  form 

(      ^/V 

r  +  wl)        y 


=  i. 


Theoretically,  therefore,  the  beam,  in  elevation,  is  the  area 
ACD,  the  curve  CAD  being  an  hyperbola  having  its  centre  at 

I  W  \ 

H  ^where  AH =  -yn-l),  and  semi-axes  equal  to 


U    and 


v  IML 

V  w 


The  maximum  depth  CD  =  A.     p\WJ-\-  W~\  =  2BC. 


362  THEORY  OF  STRUCTURES. 

A  cantilever  of  approximately  uniform  strength  may  be  ob- 
tained by  taking  the  tangents  CE,  DF  as  the  upper  and  lower 
edges  of  the  beam  instead  of  the  curves  CA,  DA.  It  may  be 

W 
easily  shown  that  the  depth  of  this  beam  at  A  is 


and  this  will  give  sufficient  sectional  area  at  A  to  bear  the 
shear  due  to  Wr 

Note. — The  dotted  lines  show  the  cantilever  of   uniform 
strength,  when  the  lower  edge  is  the  line  AB. 

Ex.  4.  A  beam  AB  supported  at  A  and  B,  and  carrying  a 

load  W^  at  the  middle  point  O. 
•\  At  a  distance  x  from  0, 

G    /^^ 1 JX    E 


Theoretically,  therefore,  the 
beam,  in  elevation,  is  the  area 

ACBD,  the  curves  CAD,  CBD  being  two  equal  parabolas, 
having  their  vertices  at  A  and  B,  respectively,  and  having 
parameters  equal  to  \p  W^ . 

The  maximum  depth  =  CD  ~  2CO  =  %  \f~pWJ . 

A  beam  of  approximately  uniform  strength  may  be  ob- 
tained by  taking  the  tangents  CE,  CG  as  the  upper  edges 
instead  of  the  curves  CA,  CB,  and  the  tangents  DFt  DH  as 
the  lower  edges  instead  of  the  curves  DA,  DB. 

The  depth  of  the  beam  at  A  and  B  is  now  EF  =  GH  —  — 

2     ' 

and  this  depth  will  give  a  sectional  area  at  the  ends  of  the  beam 
sufficient  to  bear  the  shears  at  these  point,  viz.,  — - . 

Note.— The  dotted  lines  show  the  beam  of  uniform  strength 
when  the  line  AB  is  the  lower  edge. 

Ex.  5.  A  beam  AB  supported  at  A  and  B,  and  carrying  a 
uniformly  distributed  load  W^ 


BEAMS   OF   UNIFORM  STRENGTH. 


363 


At  a  distance  x  from  the 
middle  point  O, 


This  equation  may  be  writ-  D 

^,       c  FIG.  282. 

ten  in  the  form 

*°4    _/_ 

1     pwj 

4  8 

Theoretically,  therefore,  the  beam,  in  elevation,  is  an  ellipse 
ACBD,  having  its  centre  at  O  and  axes 


AB  =  /    and     CD 


IpWJ 

=v~ 


The  maximum  depth  is  of  course  the  axis  CD  =  2 CO. 
Practically,  the  beam  must  have  a  certain  depth  at  A  and 
B  in  order  to  bear  the  shears  due  to  the  reactions  at  these 

W 
points,  viz., — -.     If  the  horizontal  tangents  at  C  and  at  D  are 

substituted  for  the  curves,  the  volume  of  the  new  beam  is  to 
the  volume  of  the  elliptic  beam  in  the  ratio  of  4  to  n. 

Note. — The  dotted  line  shows  the  beam  of  uniform  strength 
when  its  lower  edge  is  the  line  AB. 

Ex.  6.  A  beam  AB  supported  at  A  and  B,  and  carrying  a 
load  W}  at  the  middle  point  O  and  also  a  uniformly  distributed 
load  W,. 

At  a  distance  x  from  (9, 


This  equation  may  be  written  in  the  form 
.    i   WJ^ 


-;#•*. 


pi 


=  I 


w- 


THEORY  OF  STRUCTURES. 


Theoretically,  therefore,  the 
beam,  in  elevation,  is  the  area 
ACBD,  the  curves  CAD  and 
CBD  being  the  arcs  of  ellipses 
having  the  centres  at  the  points 
K  and  L,  respectively,  where 


(  w/       W  l\  * 
The  maximum  depth  CD  =  20C  =  J>*  ]  —  '-  +  -~  \  . 

(2  o      ) 

A  beam  of  approximately  uniform  strength  may  be  obtained 
by  taking  as  the  upper  edge  the  tangents  to  the  curves  at  C, 
and  as  the  lower  edge  the  tangents  to  the  curves  at  D. 

It  may  be  easily  shown  that  the  depth  at  the  ends^4  and  B 

W  A-  W 
is  now  CD       '    '        '  and  this  depth  will  make  allowance  for 

2  W  \  -\-    KKa 

W  _L  w 
the  shear  —  —  •  -  a  at  these  points. 

Note.  —  The  dotted  lines  show  the  beam  of  uniform  strength 
when  the  lower  edge  is  the  line  AB. 

CASE  b.  Assume  that  the  ratio  of  the  breadth  (b)  to  the 
depth  (y)  is  constant,  i.e.,  that  transverse  sections  are  similar. 

y  oc  b  a  ty  M, 

or  the  ordinates  of  the  profile  of  the  beam  both  in  plan  and 
elevation  are  proportional  to  the  cube  roots  of  the  ordinates 
of  the  curve  of  bending  moments. 

For  concentrated  loads  the  bounding  curves  are  evidently 
cubical  parabolas. 

CASE  c.  Assume  that  the  depth  y  is  constant.     Then 

b  a  Mt 

so  that  the  ordinates  of  the  beam  in  plan  are  directly  propor- 
tional to  the  ordinates  of  the  curve  of  bending  moments. 

CASE  d.   Assume  that  the   sectional  area  yb  is   constant. 

Then 

y  a  'M, 


FLANGED    GIRDERS,  ETC,  365 

and  the  ordinates  in  elevation  are  directly  proportional  to  the 
ordinates  of  the  curve  of  bending  moments. 

In  this  beam,  the  distribution  of  the  material  is  very  de- 
fective, as  the  breadth  b  (  =  -J  -)  must  be  infinite  when^  =  o, 

i.e.,  at  the  points  at  which  the  bending  moment  is  nil. 

Timber  beams  of  uniform  strength  are  uncommon,  as  there 
is  no  economy  in  their  use,  the  portions  removed  to  bring  the 
beam  to  the  necessary  form  being  of  no  practical  value. 

6.  Flanged  Girders,  etc. — Beams  subjected  to  forces,  of 
which  the  lines  of  action  are  at  right  angles  to  the  direction  of 
their  length,  are  usually  termed  Girders;  a  Semi-girder,  or 
Cantilever,  is  a  girder  with  one  end  fixed  and  the  other  free. 

It  has  been  shown  that  the  stress  in  the  different  layers  of 
a  beam  increases  with  the  distance  from  the  neutral  surface,  so 
that  the  most  effective  distribution  of  the  material  is  made  by 
withdrawing  it  from  the  neighborhood  of  the  neutral  surface 
and  concentrating  it  in  those  parts  which  are  liable  to  be  more 
severely  strained.  This  consideration  has  led  to  the  introduction 
of  Flanged  Girders,  i.e.,  girders  consisting  of  one  or  two  flanges 
(or  tables),  united  to  one  or  two  webs,  and  designated  Single- 
webbed  or  Double-webbed  ( Tubular]  accordingly. 


7 
( 


FIG.  284.  FIG.  285. 


T 
j 


IT      T 
i       JL 


FIG.  286.          FIG.  287.  FIG.  288.  FIG.  289.  FIG.  290. 

The  web  may  be  open  like  lattice-work  (Fig.  284),  or  closed 
and  continuous  (Fig.  285). 

The  principal  sections  adopted  for  flanged  girders  are  : 

The  Tee  (Figs.  286  and  287),  the  I  or  Double-tee  (Figs.  288 
and  289),  the  Tubular  or  Box  (Fig.  290). 

Classification  of  Flanged  Girders.  —  Generally  speaking, 
flanged  girders  may  be  divided  into  two  classes,  viz.: 


366  THEORY  OF  STRUCTURES. 

I.  Girders  witJi  Horizontal  Flanges. — In  these  the  flanges 
can  only  convey  horizontal  stresses,  and   the  shearing  force, 
which  is  vertical,  must  be  wholly  transmitted  to  the  flanges 
through  the  medium  of  the  web. 

If  the  web  is  open,  or  lattice-work,  the  flange  stresses  are 
transmitted  through  the  lattices. 

If  the  web  is  continuous,  the  distribution  of  stress,  arising 
from  the  transmission  of  the  shearing  force,  is  indeterminate, 
and  may  lie  in  certain  curves ;  but  the  stress  at  every  point  is 
resolvable  into  vertical  and  horizontal  components.  Thus,  the 
portion  of  the  web  adjoining  the  flanges  bears  a  part  of  the 
horizontal  stresses,  and  aids  the  flanges  to  an  extent  depend- 
ent upon  its  thickness. 

With  a  thin  web  this  aid  is  so  trifling  in  amount  that  it 
may  be  disregarded  without  serious  error. 

II.  Girders  with  one  or  both  Flanges  Curved. — In  these  the 
shearing  stress  is  borne  in  part  by  the  flanges,  so  that  the  web 
has  less  duty  to  perform  and  requires  a  proportionately  less 
sectional  area. 

Equilibrium  of  Flanged  Girders. — AB  is  a  girder  in  equi- 
o  librium  under  the  action  of  external 

forces,  and  has  its  upper  flange  com- 
pressed   and    its   lower   flange    ex- 


tended.    Suppose  the  girder  to  be 
FlG-  89X-  divided   into   two  segments   by  an 

imaginary  vertical  plane  MN.  Consider  the  segment  AMN. 
It  is  kept  in  equilibrium  by  the  external  forces  on  the  left  of 
MN,  by  the  compressive  flange  stress  at  N  (  =  C),  by  the 
tensile  flange  stress  at  M  (  —  T),  and  by  the  vertical  and 
horizontal  web  stresses  along  MN.  The  horizontal  web 
stresses  may  be  neglected  if  the  web  is  thin,  while  the  vertical 
web  stresses  pass  through  M  and  N,  and  consequently  have  no 
moments  about  these  points. 

Let  d  be  the  effective  depth  of  the  girder,  i.e.,  the  distance 
between  the  points  of  application  of  the  resultant  flange  stresses 
in  the  plane  MN. 

Take  moments  about  Jfand  //successively.     Then 

Cd  =  the  algebraic  sum  of  the  moments  about  M  of 


FLANGED    GIRDERS,  ETC.  367 

the  external  forces  upon  AMN  •=.  the  bending  moment  at 
MN=M. 

So,  Td  =  M\  .'.  Cd=M=  Td,  and  C  =  T. 

Hence,  the  flange  stresses  at  any  vertical  section  of  a  girder 
are  equal  in  magnitude  but  opposite  in  kind.  The  flange 
stress,  whether  compressive  or  tensile,  will  be  denoted  by  F. 

EXAMPLE.  —  A  flanged  girder,  of  which  the  effective  depth 
is  10  ft.,  rests  upon  two  supports  80  ft.  apart,  and  carries  a  uni- 
formly distributed  load  of  2500  Ibs.  per  lineal  foot.  Determine 
the  flange  stress  at  10  ft.  from  the  end,  and  find  the  area  of 
the  flange  at  this  point,  so  that  the  unit  stress  in  the  metal 
may  not  exceed  10,000  Ibs.  per  square  inch. 

The  vertical  reaction  at  each  support 

80  X  2500 
=  --  -  —  =  100,000  Ibs. 

/.  F.  10  =  M—  looooo  X  10  —  2500  X  10  X  5  =  875,000  ft.-lbs. 
,:F  =  87,500  Ibs. 

87500 

The  required  area  =  ---  =  8.75  sq.  in. 
i  oooo 

Cor.  i.  Fd=M=^I  =  ^L 
R         y 

Cor.  2.  At  any  vertical  section  of  a  girder, 
let  #,,#„,  be  the  sectional  areas  of  the  lower  and  upper  flanges, 

respectively  ; 

fuft,  be  the  unit  stresses  in  the  lower  and  upper  flanges, 
respectively.     Then 


and  the  sectional  areas  are  inversely  proportional  to  the  unit 
stresses. 

This  assumes  that  F  is  uniformly  distributed  over  the 
areas  alta9,  so  that  the  effective  depth  is  the  vertical  distance 
between  centres  of  gravity  of  these  areas.  Thus,  the  flange 
stresses  at  the  centres  of  gravity  are  taken  to  be  equal  to  the 


3^8  THEORY  OF  STRUCTURES. 

maximum  stresses,  and  the  resistance  offered  by  the  web  to 
bending  is  disregarded.  The  error  due  to  the  former  may 
become  of  importance,  and  it  may  be  found  advisable  to  make 
the  effective  depth  a  geometric  mean  between  the  depths  from 
outside  to  outside  and  from  inside  to  inside  of  the  flanges. 

Thus,  if  these  latter  depths  are  h^  ,  /^  ,  the  effective  depth 
=  Vk&  (Art.  7). 

EXAMPLE  I.  At  a  given  vertical  section  of  a  flanged  girder 
the  sectional  area  of  the  top  flange  is  10  sq.  in.,  and  the  cor- 
responding unit  stress  is  8000  Ibs.  per  square  inch.  Find  the 
sectional  area  of  the  lower  flange,  so  that  the  unit  stress  in  it 
may  not  exceed  10,000  Ibs.  per  square  inch. 
at  .  10000  ='F=  10  .  8000  ;  /.  al  =  8  sq.  in.  and  F  =  80,000  Ibs. 

Ex.  2.  A  wrought-iron  girder  weighing  w  Ibs.  per  lineal 
ft.,  of  /  ft.  span  and  d  ft.  depth,  has  horizontal  flanges  and 
a  uniform  cross-section.  The  weight  of  the  web  is  equal  to  the 
weight  of  the  flanges.  Show  that  if  the  coefficient  of  strength 
is  9000  Ibs.  per  square  inch,  the  limiting  value  of  /  is  5400^  ft., 
k  being  the  ratio  of  depth  to  span. 

wr 

Maximum  flange  stress  —  --,-  ; 


Area  of  each  flange        =  --  5-1  in.  ; 

9000  .  Sa 


and 


4wT 
Total  sectional  area        = ^  in., 

total  volume  of  girder  in  feet  =  — '• 5--;' 

9000.  8^.  144 

Hence, 

4wr  .480 

wl  =  total  WeIght  =  -95^-w:— , 

and 

d 
I  =  5400^-  =  5400^. 


FLANGED    GIRDERS,  ETC.  369 

Note.  —  The  compressive  strength  of  cast-iron  is  almost  six 
times  as  great  as  the  tensile  strength,  and  therefore  the  area 
of  the  tension  flange  of  a  girder  of  this  material  should  be 
about  six  times  that  of  the  compression  flange.  Considering, 
however,  the  difficulty  there  is  in  obtaining  sound  castings, 
and  also  the  necessity  to  provide  sufficient  lateral  strength,  it 
by  no  means  follows,  nor  is  it  even  probable,  that  the  ratio  of 
ultimate  strengths  is  the  best  for  the  working  strengths.  Some 
authorities  are  of  the  opinion  that  girders  should  be  designed 
with  a  view  to  their  elastic  strength,  and  that  therefore  the 
working  unit  stresses  in  the  case  of  wrought-iron  and  steel 
should  be  equal,  if  this  will  insure  sufficient  lateral  stability, 
and  in  the  ratio  of  2  to  I  or  3  to  I  for  cast-iron,  which  will  give 
sufficient  lateral  stability  and  make  allowance  for  defective 
castings. 

The  formula  W  =  C—r  is  often  employed  to  determine  the 

strength  of  a  cast-  or  wrought-iron  girder  which  rests  upon  twox 
supports  /  inches  apart,  d  being  its  depth  in  inches,  and  a  the 
net  sectional  area  of  the  bottom  flange  in  square  inches.  C  is 
a  constant  to  be  determined  by  experiment.  Its  average  value 
for  cast-iron  is  24  or  26,  according  as  the  girder  is  cast  on  its 
side  or  with  its  bottom  flange  upwards.  An  average  value  of 
C  for  wr  -ought-iron  is  80. 

Cor.  3.  A  girder  with  horizontal  flanges,  of  length  /  and 
depth  d,  rests  upon  two  supports,  and  is  uniformly  loaded  with 
a  weight  w  per  unit  of  length. 

The  bending  moment  at  a  vertical  plane  distant  x  from  the 
centre  is 

.      will         \  I         \i//         \       wl 


Also,  M  —  Fd  —  afd,  a  being  the  sectional  area  of  either 
flange  at  the  plane  under  consideration,  and  f  the  correspond- 
ing unit  stress. 


wrt      AX*\ 

=  —  (i--). 


3/O  THEORY  OF  STRUCTURES, 

Let  A  be  the  flange  sectional  area  at  the  centre.     Then 


Hence 


an  expression  from  which  the  flange  sectional  area  at  any  point 
of  the  girder  may  be  obtained  when  the  area  at  the  centre  is 
known. 

Cor.  4.  F  represents  indifferently  the  sum  of  the  horizontal 
elastic  forces  either  above  or  below  the  neutral  axis,  and  is 
therefore  proportional  to  A,  the  sectional  area  of  the  girder ; 
d  is  the  distance  between  the  centres  of  resultant  stress  and  is 
proportional  to  D,  the  depth  of  the  girder. 

/.  Mat  AD  =  CAD, 

a  form  frequently  adopted  for  solid  rectangular  or  round  gird- 
ers, but  also  applicable  to  other  forms. 

Remark. — The  effective  length  of  a  girder  may  be  taken  to 
be  the  distance  from  centre  to  centre  of  bearings. 

The  effective  depth  depends  in  part  upon  the  character  of 
the  web,  but  in  the  calculation  of  flange  stresses  the  following 
approximate  rules  are  sufficiently  accurate  for  practical  pur- 
poses : 

If  the  web  is  continuous  and  very  thin,  the  effective  depth 
is  the  full  depth  of  the  girder. 

If  the  web  is  continuous  and  too  thick  to  be  neglected,  the 
effective  depth  is  the  distance  between  the  inner  surfaces  of 
the  flanges. 

If  the  web  is  open  or  lattice-work,  the  effective  depth  is  the 
vertical  distance  between  the  points  of  attachment  of  the 
lattices. 

If  the  flanges  are  cellular,  the  effective  depth  is  the  distance 
between  the  centres  of  the  upper  and  lower  cells. 


EXAMPLES   OF  MOMENTS   OF  INERTIA.  $Jl 

7.  Examples  of  Moments  of  Inertia. — (a)  Double-tee  Sec- 
tion.— First,  suppose  the  web  to  be  so  thin  that 
it  may  be  disregarded  without  sensible  error.          ""T"^ 

K2' 

Let  the  neutral  axis  pass  through  G,  the  cen-    -k 
tre  of  gravity  of  the  section.  J1 


Let  a^ ,  a^  be  the  sectional  areas  of  the  lower         A 
and  upper  flanges,  respectively,  and  assume  that         FlG' 2Q2' 
each  flange  is  concentrated  at  its  centre  line. 

Let  hl ,  hz  be  the  distances  of  these  centre  lines  from  G. 

Let  //,  +  h,  =  d. 

Approximately,  /  =  aji*  -\-  aji*. 

Also,      (al  +  a<i)/il  =  a^d,     and     (al  +  #2)/z2  =  a^. 


•'•  I=a\^aJ+a* 

Again,  if/,,  /2  are,  respectively,  the  unit  stresses  in  the 
metal  of  the  lower  and  upper  flanges, 

M  =  j-I  —  f&d,     and  also     =  -—/  =  f^a^d. 

If  a,  =  a9  =  a,  /  =  /,=/,  suppose,  and  M  =  fad. 
Second.  Let  the  web  be  too  thick  to  be  neglected. 
As  before,  let  the  neutral  axis  pass  through  £,  the  centre 
of  gravity  of  the  section. 

Let  al ,  #2  be   the  sectional  areas  of  the  lower 
and   upper  flanges,  respectively,   and   assume  that 
n   each  flange  is  concentrated  at  its  centre  line. 
FIG.  29*a.  Let  ^  ^  ^  be  the  sectjonai  areas  of  the  portions 

of  the  web  below  and  above  G,  respectively. 

Let  h^ ,  hi  be  the  distances  from  G  of  the  lower  and  upper 
flange  centre  lines. 

Let  hl  -f  ^2  =  d. 

Approximately, 


3/2  THEORY  OF  STRUCTURES. 

Also,  fa,  +  —  j^,  =  (a^  +  —  U8,  and  this  equation,  together 


with  hl  +  hi  =  d,  will  give  the  values  of  ^  ,  &,  ;    hence  the 
value  of  /  may  be  determined. 

As  before,  £/  =  J/=/V. 


A' 
Let  #,  =  #3  =  A  and  #3  =  #4  =  —  .    Then 


Hence, 


/being  the  unit  stress  in  either  flange. 

Thus,  the  web  aids  the  girder  to  an  extent  equivalent  to  the 
increase  which  would  be  derived  by  adding  one-sixth  of  the 
web  area  to  each  flange.  If  the  weight  of  the  material  remains 
constant,  M  increases  with*/.  At  the  same  time  the  thickness 
of  the  web  diminishes,  its  minimum  value  being  limited  by  cer- 
tain practical  considerations  (Art.  8).  Hence  it  follows  that 
the  distribution  of  material  is  most  effective  when  it  is  concen- 
trated as  far  as  possible  from  the  neutral  axis  (Art.  5). 

N.B.  —  It  must  be  remembered  that  fl  and  /a  are  not  the 
maximum  stresses.  If  /,  ,  /a  are  the  thicknesses  of  the  lower 
and  upper  flanges,  respectively,  then 


and 


maximum  tension  =       ^—, - 


A  +  K 

maximum  compression  =/2- — -, . 

fin 


EXAMPLES   OF  MOMENTS   OF  INERTIA.  3/3 

Again,  take  moments  about  G.     Then 


or 


which  gives  a  relation  between  the  flange  and  web  areas  if 
/j  ,  /,  are  known. 

For  example,  take  ft  =  2^/j.     Then 


a  formula  which  agrees  very  closely  with  modern  practice  in 
cast-iron  girders. 


The  principles  of  construction  require  a  beam  or  girder 
to  be  designed  in  such  a  manner  as  to  be  of  uniform 
strength,  i.e.,  equally  strained  at  every  point.  An  exception, 
however,  is  usually  made  in  the  case  of  timber  beams  or  girders. 
The  fibres  of  this  material  are  real  fibres  and  offer  the  most 
effective  resistance  in  the  direction  of  their  length,  so  that  if 
they  are  cut,  their  remaining  strength  is  due  only  to  cohesion 
with  the  surrounding  material.  Besides,  there  is  no  economy 
to  be  gained  by  removing  a  lateral  portion,  as  the  waste  is  of 
little,  if  any,  practical  value. 

EXAMPLE.  The  lower  and  upper  flanges  of  the  section  of  a 
girder  are  I  in.  and  I  J-  in.  thick,  respectively,  and  are  each  24  in. 
wide  ;  the  effective  depth  of  the  girder  is  48  in.,  and  the  web 
is£  in.  thick.  Determine  the  position  of  the  neutral  axis  ;  also 
find  the  flange  unit  stresses  when  the  bending  moment  at  the 
given  section  is  250  ft.-tons.  Using  the  preceding  notation, 

al  =  24  sq.  in.,    az  =  36  sq.  in.,    and    a3  -f-  #4  =  24  sq.  in. 

The  centre  of  gravity  of  the  web  is  half-way  between  AB 
and  CD.  Thus, 

24/Z,  +  24(//,  -  24)  =  36(48  -  //,), 


374    -  THEORY  OF  STRUCTURES. 

or 

_//  \AA>f 

hi  =  —  —      and     h^  —  ---  ,  defining  the  position  of  G. 

Again, 

192    i       96  144    i        72 

a  =  —  -  .  —  =  —  sq.  in.      and     a.  =  --  .  -  =  —  sq.  in. 
727  7     2        7    4 


Also, 


M  =  250  ft.-tons  =  3000  inch-tons. 


.-.  /,  =  2-g-9^-  tons  per  sq.  in.     and    ft  =  i||-f  tons  per  sq.  in. 

Third.  It  is  often  convenient  to  calculate  the  moment  of 
inertia  of  a  built  beam  symmetrical  with  respect  to  the  neutral 
axis,  as  follows : 

Let  Fig.  293  represent  the  section  of  such  a  beam,  com- 
posed of  equal  flanges  connected  with  the  web  by  four  equal 
angle-irons. 

Let  the  width  AF  oi  the  flange  =  a. 
u     the  side  BC(=  DE)  of  an  angle-iron  =  b. 
"     thickness  GH(=  KL)  of  an  angle-iron  =f. 


EF  _ 


,  be  the  outside  depth  of  the  section. 
,  "     "     depth  between  flanges. 


FIG.  293.  «     h^  " 

Let  ^3  be  the  depth  between  the  faces  MN,  M'N'. 
"     h,  "     "         "  "  "       "      ATZ,  A'7/:7. 


EXAMPLES  OF  MOMENTS   OF  INERTIA. 


375 


In  this  value  of  /,  the  weakening  effect  due  to  the  rivet- 
holes  in  the  tension  flange  has  been  disregarded.*  If  it  is  to  be 
taken  into  account,  let/  be  the  diameter  of  the  rivets. 

The  centre  of  gravity  of  the  section  is  now  moved  towards 
the  compression  flange  from  its  original  position  through  a 
distance 


and  the  moment  of  inertia  of  the  net  section  with  respect  to 
the  axis  through  the  new  C.  of  G.  is 


A'  being  the  net  area  of  the  section,  and  /  having  the  value 
given  above. 

Fourth.  The  value  of  7  for  a  double-tee  section   may  be 
more  accurately  determined  as  follows  : 

Let  the  area  of  the  top  flange  be  A1  ,  and 
its  depth  hr 

Let  the  area  of  the  bottom  flange  be  A^  , 
and  its  depth  //8. 

Let  the  area  of  the  web  flange  be  A^  ,  and 
its  depth  /z2. 

Let  A,-\-  Af{-  A3  =  A,  and  /z,+  //„+  hz=k.  FIG.  294. 

Let  G   be  the  centre  of  gravity  of  the  section. 
"     Gl      "  "  "  "      top  flange.  m 

"      web. 
"      bottom  of  flange. 

Let  yv  be  the   distance  of  G  from   the  upper  edge  of  the 
section. 

Let  yt  be  the  distance  of  G  from  the  lower  edge  of  the 
section. 

Take  moments  about  Gs.     Then 


THEORY  OF  STRUCTURES. 
or 

GG    =  *,(*,  +  2*.  +  *J  +  *.(*,  +  *. 

2,A 

So, 

rr        -A^+ 

^  ~ 
and 


Hence, 

h      k        h 


2A 


2A 

+  kj  -  Afa  +  ^3)  -  A^  -  h 


2 

So, 


=  GG,  +  ~=  etc. 


Again,  /,  with  respect  to  G, 


G,ff+A,.  G,ff  +  AS.  G,G', 

A&*  +  A,&,*  +  A,A* 
/,  being  equal  to     '  '       -    '  '  ~  —  —  . 

Hence, 

7  =  7'  +  ^'  + 


\A^  +  2A,  +  A.)  +  A,(A,  +  /,,)'. 


EXAMPLES  OF  MOMENTS  OF  INERTIA.  ^77 

A1A^1+^+2AlAtA,(/l1+/l^1+2^+^ 


+  A,A,'(At  +  ky  -  2AlA^i,(A,  +  A,)(A,  +  A,) 
+  A,A,'(A,  +  A,)'  +  A,A'(A,  +  2A,  +  A,)' 
+  2A1A,A,(A1  +  2*.  +  h,)(h,  +  A,) 


1+  2A1A,A3(/i1  +  2/1,  +  h^(h,  +  h,  +  h,  +  h,) 
A.A^  +  k$A  +  A,A,(/i,  +  k$A  1 


+  (4&  +  AfAJfa  +  2h,  +  h$ 

W*  +  WA 
-  A.A^/^  +  2//a  +  h$A 

Hence,  finally, 


12 


Cor.  I.  If  7/t  and  ht  are  small  compared  with  ^,,  put 


Then 


y  = 


2A 


h, 
=  --  — ,  nearly, 


THEORY  OF  STRUCTURES. 

and 

A  A'  +  A.  (h'  -  k-^^]  +  AM 


12 


,,^A,    ,  A,A,+A,A, 
tttT"  ~4A 


4A 

Note.  —  If  At  is  also  very  small,  as  in  the  case  of  an  open 
web,  then 

A  A  A 

i/2  =  h'  —  j-     and     /  =  h'  2  —  j-2,  approximately. 

si.  si. 

Cor.  2.  Let  ya,  yb  be  the  distances  of  G  from  the  upper  and 
lower  edges,  respectively  ;  let  fa  ,  fb  be  the  corresponding 
maximum  working  unit  stresses. 

From  the  preceding  corollary,  yb  =  y^  =  —  -  -  —  —  -  —  -, 

2  2/i 

or 

A.  +  A.  +  A,  __  ya+yb 
2A.  +  A,  2yb 


3  l  yb          *     2%  1  fb 

Hence, 


EXAMPLES   OF  MOMENTS   OF  INERTIA. 


379 


and 


,,,  1, 

~lt          12  + 


4A 


12A 


A  +  A:  +  4(A,  + 


+  A, 


6  /.  +/. 

r&, 


Fifth.  T-section. 

Let  the  area  of  the  flange  be  A, ,  and  its 
depth  h,. 

Let  the  area  of  the  web  be  A9 ,  and  its 
depth  h^. 

LetAi  +  A,  =  A,     and     h,  +  h^  =  h. 

Let  G  be  the  centre  of  gravity  of  the  sec- 
tion, G^  of  the  flange,  and  G9  of  the  web. 


FIG. 


295- 


Let  yl  be  the  distance  of  G  from  foot  of  the  web. 
Then 


.l  - 

2  2 


3^0  THEORY  OF  STRUCTURES. 

and 

_  ^i  +  *«  I  AA  -  A  A  =  *i  AA  -  AA 

2  2(Al+A,)          2   '  2A 

Again,  ,. 

^  ^      ^ii/  AJi          .      ~  ~  h.       AJi 

GlG  =  ^  +  k,-yt=^     and     G.G  =*—;-  =  £.. 

Hence  /,  with  respect  to  a  horizontal  line  through  G, 

=  A^+A,.  G>G'+A/£  +  A..G.G;    '' 

which  reduces  to 


. 
12  4A 

Cor.  I.  If  h^  is  very  small  as  compared  with  ^a,  put 


=  AJf  +  A.(?-  -)  =  (A,  +     )  k>,  nearly, 


then 
or 

and 

1  = 

or 


12    '     4A  r 

Cor.  2.  Let  ya  be  the  distance  of  the  compressed,  or  upper, 
side  from  the  neutral  axis. 


12  4A 


TO  DESIGN  A    GIRDER    OF    UNIFORM  STRENGTH.      381 

Let  yb  be  the  distance  of  the  stretched,  or  lower,  side  from 
the  neutral  axis. 

Let/a  be  the  crushing  unit  stress,  fb  the  tensile  unit  stress. 

h'  2.A     I   A. 
From  the  preceding,  ya  —  -    '-  —  ^—-  —  -  ;   but  h'  =  ya  -\-yb  ; 


m>,      ^          =  >  = 

2        A,+At  2yb  2fb 


Hence,  /becomes 


2~7  —  i  —  7"'  —  7 

fa+fb  fa 


Note.  —  Although  the  preceding  approximate  methods  are 
often  useful,  they  can  only  be  regarded  as  tentative  and  should 
always  be  checked  by  an  accurate  determination  of  the  moment 
of  inertia  and  of  the  position  of  the  neutral  axis. 

8.  To  design  a  Girder  of  Uniform  Strength,  of  an 
I-section  with  equal  Flange  Areas,  to  carry  a  Given 
Load. 

Let  y  be  the  depth  of  the  girder  at  a  distance  x  from  its 
middle  point. 

Let  A  be  the  sectional  area  of  each  flange  at  a  distance  x 
from  its  middle  point. 

Let  A'  be  the  sectional  area  of  the  web  at  a  distance  x  from 
its  middle  point. 

Let  M  be  the  bending  moment  at  a  distance  x  from  its  mid- 
dle point. 

Let  5  be  the  shearing  force  at  a  distance  x  from  its  middle 
point.  Then  * 


/being  the  safe  unit  stress  in  tension  or  compression. 


382  THEORY  OF  STRUCTURES. 

Web.  —  Assume  that  the  web  transmits  the  whole  of  the 
shearing  force.  This  is  not  strictly  correct  if  the  flange  is 
curved,  as  the  flange  then  bears  a  portion  of  the  shearing  force. 
The  error,  however,  is  on  the  safe  side. 

Theoretically,  the  web  should  contain  no  more  material  than 
is  absolutely  necessary. 

Let  fs  be  the  safe  unit  stress  in  shear.     Then 

A'-S 
~ 


and  the  sectional  area  is,  therefore,  independent  of  the  depth. 

A.'        S 
The  thickness  of  the  web  =  —  =  -7—, 


but  this  is  often  too  small  to  be  of  any  practical  use. 

Experience  indicates  that  the  minimum  thickness  of  a  plate 
which  has  to  stand  ordinary  wear  and  tear  is  about  J  or  T5^-  in., 
while  if  subjected  to  saline  influence  its  thickness  should  be 
•f  or  |-  in.  Thus,  the  weight  of  the  web  rapidly  increases  with 
the  depth,  and  the  greatest  economy  will  be  realized  for  a  cer- 
tain definite  ratio  of  the  depth  to  the  span. 

The  thickness  of  the  web  in  a  cast-iron  girder  usually 
varies  from  I  to  2  in. 

In  the  case  of  riveted  girders  with  plate  webs  of  medium 
size,  all  practical  requirements  are  effectively  met  by  specifying 
that  the  shearing  stress  is  not  to  exceed  one-half  of  the  flange 
tensile  stress,  and  that  stiffeners  are  to  be  introduced  at  inter- 
vals not  exceeding  twice  the  depth  of  the  girder  when  the 
thickness  of  the  web  is  less  than  one-eightieth  of  the  depth. 
Again,  it  is  a  common  practical  rule  to  stiffen  the  web  of  a 
plate  girder  at  intervals  approximately  equal  to  the  depth  of 
the  girder,  whenever  the  shearing  stress  in  pounds  per  square 

Crra    \ 
I  -\  —  '•    —  j,  H  being  the  ratio  of  the 

depth  of  the  web  to  its  thickness. 

Flanges.  —  First.  Assume  that  the  flanges  have  the  same 
sectional  area  from  end  to  end  of  girder. 


TO  DESIGN  A    GIRDER   OF   UNIFORM  STRENGTH.      383 

If  the  effect  of  the  web  is  neglected, 

M 

and  the  depth  of  the  beam  at  any  point  is  proportional  to  the 
ordinate  of  the  bending-moment  curve  at  the  same  point. 

For  example,  let  the  load  be  uniformly  distributed  and  of 
intensity  w ;  and  let  /  be  the  span.     Then 


FIG.  296. 

and  the  beam  in  elevation  is  the  parabola  ACJ3,  having   its 
vertex  at  C  and  a  central  depth  CO  =  TTT">     The  depths  thus 

determined  are  a  little  greater  than  the  depths  more  correctly 
given  by  the  equation 

M 

y  = 


Second.  Assume  that  the  depth  y  of  the  girder  is  constant. 
Then 

A1       M 


and,  neglecting  the  effect  of  the  web,  the  area  of  the  flange  at 
any  point  is  proportional  to  the  ordinate  of  the  curve  of  bend- 
ing moments  at  the  same  point. 

Let  the  load  be  uniformly  distributed  and  of  intensity  w  ; 
also,  let  the  flange  be  of  the  same  uniform  width  b  throughout. 


A   1     2      3  0  5     6      7    B 

FIG.  297, 

The  flange,  in   elevation,  is  then  the  parabola  ACB,  having 

its  vertex  at   C  and  its  central  thickness  CO  —  ^7-7.      Such 

Zfyb 


THEORY  OF  STRUCTURES. 

beams  are  usually  of  wrought-iron  or  steel,  and  are  built  up  by 
means  of  plates.  It  is  impracticable  to  cut  these  plates  in  such 
a  manner  as  to  make  the  curved  boundary  of  the  flange  a  true 
parabola  (or  any  other  curve).  Hence,  the  flange  is  generally 
constructed  as  follows : 

Draw  the  curve  of  bending  moments  to  any  given  scale. 
By  altering  the  scale,  the  ordinates  of  the  same  curve  will 
represent  the  flange  thicknesses.  Divide  the  span  into  seg- 
ments of  suitable  lengths. 

From  A  to  I  and  B  to  7  the  thickness  of  the  flange  is 
\a  —  7/;  from  I  to  2  and  7  to  6  the  thickness  is  2b  =  6e ;  from 
2  to  3  and  6  to  5  the  thickness  is  $c  =  ^d\  and  from  3  to  5  the 
thickness  is  CO. 

The   more   correct  value  of  A(  =— -p-1    is   somewhat 

V     fy      67 

less  than  that  now  determined,  but  the  error  is  on  the  safe 
side. 

Again,  at  any  section, 

E      2f 

-£  —  — ,      and  hence      R  oc  y,  the  depth. 

Thus  the  curvature  diminishes  as  the  depth  increases,  so 
that  a  girder  with  horizontal  flanges  is  superior  in  point  of 
stiffness  to  one  of  the  parabolic  form.  The  amount  of  metal 
in  the  web  of  the  latter  is  much  less  than  in  that  of  the 
former.  If  great  flexibility  is  required,  as  in  certain  dyna- 
mometers, the  parabolic  form  is  of  course  the  best. 

9.  Deflection  of  Girders. — The  principles  of  economy  and 
strength  require  a  girder  to  be  designed  in  such  a  manner  that 
every  part  of  it  is  proportioned  to  the  greatest  stress  to  which 
it  may  be  subjected.  When  such  a  girder  is  acted  upon  by 
external  forces,  it  is  uniformly  strained  throughout,  and  in 
bending,  the  neutral  axis  must  necessarily  assume  the  form  of 
an  arc  of  a  circle,  provided  the  limit  of  elasticity  is  not  ex- 
ceeded. It  might  be  supposed  that  the  curve  of  deflection  is 
dependent  upon  the  character  of  the  web,  and  this  is  doubtless 
the  case,  but  experiments  indicate  that  so  long  as  the  flange 


DEFLECTION  OF  GIRDERS.  385 

unit  stresses  are  unaltered  in  amount,  the  influence  of  the  web 
may  be  disregarded  without  sensible  error. 

Let  f  be  the  unit  stress  in  the  beam  at  a  distance  y  from 
the  neutral  axis  ;  let  d  be  the  depth  of  the  beam.     Then 

/      M      E 

-  =  -T  =  -=.  =  a  constant, 

y        1        K 

assuming  that  the  neutral  axis  is  an  arc  of  a  circle  of  radius  R. 
But    y  oc  d,     and 

7  =  Ak*  a  Ad\ 

Hence  f  a  y  a  d\  and  if  the  depth  is  constant,  /is  also  con- 
stant and  the  beam  is  of  uniform  strength. 
If  the  area  A  is  constant, 


EXAMPLE  I.    A  cantilever  bent  under  the  action  of  exter- 
nal forces,  so  that  its  neutral  axis  AB  assumes 
the  form  of  an  arc  of  a  circle  having  its  centre 

*  O. 

Draw  the  verticals  OA,  BF,  and  the  horizon- 


The  vertical  deviation  of  B  from  the  hori- 
zontal,  viz.,  BFy   is    the    maximum    deflection.     ,'/' 
Denote  it  by  D.  6 

Let  radius  of  circle  =  R. 

Since  the  deflection  is  very  small,  BE  is  approximately 
equal  to  AB  (  =  /),  the  length  of  the  cantilever. 

/.  r  =  BE"  =  AE(2R  -  AE)  =  2RD  -  D*  =  2RD, 

as  D1  may  be  disregarded  without  much  error. 

Also,  the  deflection  at  any  point  distant  x  from  A  is  evi- 
dently ^.  If  /is  the  stress  in  the  material  at  a  distance  y 
from  the  neutral  axis, 

f__EL_^DE  2DEy 

~y  ~  R  ~   '  S  ~'  f~~    ~lf~  ' 


386 


THEORY   OF  STRUCTURES. 


Ex.  2.  A  girder  resting  upon  two  supports  at  A  and  B 
is  bent  under  the  action  of  external 
forces  so  that  its  neutral  axis  ACB 
assumes  the  form  of  an  arc  of  a  circle 
having  its  centre  at  O. 

Draw  the  vertical  OC,  meeting  the 
horizontal  AB  in  F. 

CF  is   the   maximum    deflection  ; 
denote  it  by  D. 

Since  D  is  very  small,  its  square 
may  be  disregarded  and  the  horizontal  AB  may  be  supposed 
equal  to  the  length  ACB  (  =  /)  of  the  girder,  without  much 
error.  Then 


FIG.  299. 


-  =  AF*  =  FC(zR  -  FC)  =  2RD  -  D1  =  2RD. 


Hence, 


/      & 

Also,  since  —  =  -5, 
y       R 


f~ 


WEy 


x 
The  deflection  at  a  distance  x  from  F  =  D  —  — . 

Ex.  3.  A  timber  beam  of  20  ft.  span,  is  12  in.  deep  and  6 
in.  wide :  what  uniformly  distributed  load  ( W)  will  deflect  the 
beam  I  in.,  E  being  1,200,000  Ibs.  ? 

By  Ex.  2, 

(240)2 


.'.  R  —  7200  in. 


_E  T  _  1200000 
8      '12"J^~'       7200 


12 


I20OOOO    6  .  I29 
720O  12 


.-.  W=  4800  Ibs. 


DEFLECTION   OF  GIRDERS.  387 

Ex.  4.  Let  sl ,  fl ,  dlt  and  ja ,  /2 ,  4 ,  respectively,  be  the 
length,  unit  stress,  and  distance  from  the  neutral  axis  of  the 
stretched  and  compressed  outside  fibres  in  Examples  (i)  and 

(2). 

Let  dl  -f-  d^  =  d  —  the  total  depth  of  the  girder. 
Hence,  from  similar  figures, 


d,  s,      R-d, 

-       and     7-— £- 


/  2P              ""    Z? 

*  A'-     '       •   •    A 

Also, 

.  /i+/«  IL  '""•*•  _  ^  +  ^« 


Ex.  5.  A  truss  of  span  120  ft.  and  15  ft.  deep  is  strained 
so  that  the  flange  tensile  and  compressive  unit  stresses  are 
10,000  and  8000  Ibs.,  respectively.  Find  the  deflection,  and 
difference  of  length  between  the  extreme  fibres. 


30000000  i  20 

.'.  s1  —  J3  =  .864  in.,      and      R  =  25,000  ft. 


Hence  also    D  =  =  -864  in. 


10.  Camber.  —  Owing  to  the  play  at  the  joints,  a  bridge- 
truss,  when  first  erected,  will  deflect  to  a  much  greater  extent 


388  THEORY  OF  STRUCTURES. 

than  is  indicated  by  theory,  and  the  material  of  the  truss  will 
receive  a  permanent  set,  which,  however,  will  not  prove  detri- 
mental to  the  stability  of  the  structure,  unless  it  is  increased 
by  subsequent  loads. 

If  the  chords  were  made  straight,  they  would  curve  down- 
wards, and,  although  it  does  not  necessarily  follow  that  the 
strength  of  the  truss  would  be  sensibly  impaired,  the  appear- 
ance would  not  be  pleasing. 

In  practice  it  is  usual  to  specify  that  the  truss  is  to  have 
such  a  camber,  or  upward  convexity,  that  under  ordinary  loads 
the  grade  line  will  be  true  and  straight. 

The  camber  may  be  given  to  the  truss  by  lengthening  the 
upper  or  shortening  the  lower  chord,  and  the  difference  of 
length  should  be  equally  divided  amongst  all  the  panels. 

The  lengths  of  the  web  members  in  a  cambered  truss  are 
not  the  same  as  if  the  chords  were  horizontal,  and  must  be 
carefully  calculated,  otherwise  the  several  parts  will  not  fit 
accurately  together. 

To  find  an  approximate  value  for  the  camber,  etc.  : 

Let  d  be  the  depth  of  the  truss. 

Let  sl ,  ^2  be  the  lengths  of  the  upper  and  lower  chords, 
respectively. 

Let  /! ,  /2  be  the  unit  stresses  in  the  upper  and  lower 
chords,  respectively. 

Let  dl ,  d^  be  the  distances  of  the  neutral  axis  from  the 
upper  and  lower  chords,  respectively. 

Let  R  be  the  radius  of  curvature  of  the  neutral  axis. 

Let  /  be  the  span  of  the  truss. 

d,       *.  —  //i  di       l~  **      /3 

-£  =  ±j-  =  -£      and      -£  =  — —  =  -£  ,  approximately, 

the  chords  being  assumed  to  be  circular  arcs. 

Hence,  the  excess  in  length,  of  the  upper  over  the  lower 
chord, 


STIFFNESS.  389 

Let  xl  ,  x^  be  the  cambers  of  the  upper  and  lower  chords, 
respectively.  R-\-  dl  and  R  —  d^  are  the  radii  of  the  upper 
and  lower  chords,  respectively. 

By  similar  figures,  the  horizontal  distance  between  the  ends 

r>     1        » 

of  the  upper  chord  =  —  -=  —  V,  and  the  horizontal  distance  be- 
tween the  ends  of  the  lower  chord  =  -  5~~V. 


Hence, 


R 

and 


(  j 


=  x,  .  2(R  +  O,  approximately, 


—  ^)»  approximately. 


i      i     *  \         A  I  i 

8^*        R '  oR\        Ri 

r     iy 

Hence,  approximately,  the  camber  =  —  —  —  . 

Note. — The  deflection  of  a  well-designed  and  well-built 
truss  is  often  much  less  than,  and  should  never  exceed,  I  inch 
per  100  ft.  of  span  under  the  maximum  load. 

II.  Stiffness. — If  D  is  the  maximum  deflection  of  a  girder 

W  D 

of  span  /  under  a  load    W,  then  — -,  or  more  usually  —,  is  a 

measure  of  the  stiffness  of  the  girder. 

In  practice,  the  deflection  of  an  iron  or  a  steel  girder,  under 

the  working  load,  should  lie  between  — —  and  ^ — ,  i.e.,  it  is 
limited  to  i  or  2  in.  per  100  ft.  of  span,  and  rarely  exceeds 

,  or  1.2  in.  per  100  ft.  of  span. 

1000 

A  timber  beam  should  not  deflect  more  than  -^-,  or  i  in. 

360 
per  30  ft.  of  span. 


390  THEORY  OF  STRUCTURES. 

Let  Ml  be   the  bending  moment   at   the   most   deflected 
point.     Then 


Also, 


f  being  a  numerical  coefficient  (in  Art.  9,  Ex.  \,p  =  i  ;  in  Ex. 
2,/  =  i). 
Thus 

„      £I 


gives  the  bending  moment  M^  to  which  the  girder  of  a  speci- 
fied stiffness  y  may  be  subjected. 

Again,  if  the  material  is  to  bear  a  certain  specified  unit 
stress  y,  the  maximum  bending  moment  M^  to  which  the 
girder  may  be  subjected  is  given  by  the  equation 


,  =        =  -, 

c          qd 

* 

q  being  a  numerical  coefficient  less  than  unity,  depending  upon 
the  form  of  the  section. 

Cczteris  paribus,  the  ratio  of  depth  to   span  may  be  fixed 
by  making  the  stiffness  and  strength  of  equal  importance.  Then 

M  —  M       and  therefore 


•*-•*  I  ~    ]    J     r 

~pl\l)~~  qd  ' 
or 


d~\l 


DISTRIBUTION   OF  SHEARING   STRESS. 


391 


In  practice  the  proper  stiffness  of  a  girder  is  sometimes 
secured  by  requiring  the  central  depth  to  lie  between  —  and 

— ,  its  value  depending  upon  the  material  of  which  the  girder 

is  composed,  its  sectional  form,  and  the  work  to  be  done. 

EXAMPLE. — A  cast-iron  beam  of  rectangular  section  and  of 
20  ft.  span  carries  a  uniformly  distributed  load  of  20  tons ;  the 
coefficient  of  working  strength  is  2  tons  per  sq.  in. ;  the  stiff- 
ness is  .001  ;  E  is  8000  tons.  Find  the  dimensions  of  the  beam 

viz.,  b  the  breadth  and  d  the  depth. 

• 

20.20          _fr_     bd*  _bd* 
jyj.  —      !^     .12  —    j.  ^^  2  .,  •  ^^  —  ! 
8  c  63 


Also, 


.-.  bd*  =  1800. 


20 .  20  EIiD\      8 .  8000  .  bd* 

--.12  =  M=—7(-.-)  = .(.ooi); 

Pl\l I          12  .  20.  12       V 

.-.  bd*  =  27000. 


8 


Hence, 


2700O 

-f= — 

1800 


=  15  in.     and     b  =  8  in. 


12.  Distribution  of  Shearing  Stress. — Let  Figs.  300  and 

301  represent  a  slice  of  a  beam  bounded  by  two  consecutive  sec- 

A 

PX~       j         >    Q 

«-^— 4-2 •* 

0/~ 
*d    i 


B-B' 

FIG.  300. 

tions  AB,  A'B' ,  transverse  to  the  horizontal  neutral  axis  O Of . 
Let  the  abscissae  of  these  sections  with  respect  to  an  origin 


392  THEORY  OF  STRUCTURES. 

in  the  neutral  axis  be  x  and  x  -f-  dx,  so  that  the  thickness  of 
the  slice  is  dx. 

In  the  limit,  since  dx  is  indefinitely  small,  corresponding 
linear  dimensions  in  the  two  sections  are  the  same. 

Let  /  be  the  moment  of  inertia  of  the  section  AB  (or  A ' B' 
in  the  limit)  with  respect  to  the  neutral  axis. 

Let  c  be  the  distance  of  A  (or  A'  in  the  limit)  from  the 
neutral  axis. 

Let/j ,  ft  be  the  unit  stresses  at  A  and  A ,  respectively. 

Consider  the  portion  A  CCA'  of  the  slice,  CC '  being 
parallel  to  and  at  a  distance  Y  from  the  neutral  axis.  Since 
it  is  in  equilibrium,  the  algebraic  sum  of  the  horizontal  forces 
acting  upon  it  must  be  nil.  These  forces  are: 

The  total  horizontal  force  upon  ACC, 

11     A'C'C',  and 
shear  along  the  surface  CC'. 

The  horizontal  force  upon  an  element  PQ  of  thickness  dy 
and  at  a  olistance  y  from  the  neutral  axis 


2  being  the  width  PQ.     Thus  the  total  horizontal  force  upon 

"ACC 


=£  (*  yzdy  =  ^Ay, 
c  j  Y  c 


A  being  the  area  of  ACC,  and  y  the  distance  of  the  centre  of 
gravity  of  this  area  from  O0r. 

Similarly, "the  total  horizontal  force  upon  A'C'C' 


/      J         J    ~   '      .,         J' 


DISTRIBUTION   OF  SHEARING  STRESS. 


393 


//       f\    - 
Hence,  f  —  —  —-\Ay  =  difference  of  the  horizontal  forces 

upon  A CC  and  A'C'C', 
=.  horizontal  shear  along  CC' , 
=  qwdx ; 

q  being  the   intensity  of   this  shear,  and   w  the  width  of  the 
section  at-CC. 

Let  M  and  M  —  dM  be  the  bending  moments  at  the  two 
consecutive  sections  AB,  A '  B '.     Then 


M=--I    and    M-dM=--I, 
c  c 


and  therefore 


Hence, 


dM 


dM       - 


=  (---)'• 

\C         c  I 


J 

=  qwdx, 


or 


dM  Ay       S  A- 

qw  =  ~dx  T  =  7A*> 


since =  shearing  force  at  the  section  AB  =  5. 

dx 

EXAMPLE  I.  Solid  rectangular  section  of  width  b. 


12 


or 


FIG.  302. 


and  the  intensity  of  the  shear  at  any  point  of  AB  may  be  rep- 
resented by  the  horizontal  distance    of   the    point    from    the 

parabola  A  VB,  having  its  vertex  at  F,  where  OV '  —  -  — . 

be 


394  THEORY  OF  STRUCTURES. 

The  maximum  intensity  of  shear  is  at  O  and  its  value  is 

<?^  =  3-^-. 

4  be 
The  value  of  the  average  intensity  is 

_      5 
~"   b  .  2c' 

Ex.  2.    A  hollow  rectangular  section ;    B  and  2c  being  the 
external  and  B'  and  2c'  the  internal  width  and  depth. 
At  the  neutral  axis, 

g(B-B')^S-\^-l 


Thus,  as  in  Ex.  I,  the  intensity  of  shear  is  again  greatest  at 
the  neutral  plane,  i.e.,  when  Y  =  o. 

Ex.  3.  Solid  circular  section  of  radius  c. 


* 


Ay  =y2y  Vc*  -fdy  =  |(,'  -  Y* 


w  =  2c  - 


and     /  =  — 


and  the  intensity  of  the  shear  at  any  point  of  AB  may  be  rep- 
resented by  the  horizontal  distance  of  the  point  from  the  pa- 

rabola AVB,  where  OV=  -^-. 

3*** 


and 


•'•    Qmax.   '•   $av.   '   ''  4  *    3' 


DISTRIBUTION  OF  SHEARING   STRESS.  395 

Ex.  4.  A  double-flanged  section,  each  of  the  flanges  con- 
sisting of  five  8-in.  X  i-in.  plates  riveted  to  a  24-in.  X  i-in.  web 
by  two  3-in.  X  3-in.  X  i-in.  angles. 

To  find  the  intensity  of  shear  at  the  surface  of  contact 
between  the  angles  and  the  flange : 


Ay  =  20  X  I3i  =  265  ;     w  =  6J  in. ;     /  =  8975!, 


neglecting  the  effect  of  the  rivet-holes  in  the  tension  flange. 
Hence 

0  2120 

q  =  o . 

466739 


Let  5  =  49  tons.     Then  q  —  .2226  ton  per  square  inch. 
Let  the  rivets  have  a  pitch  of  4  in.,  then 

61 
the  total  shear  on  each  rivet  =  —  X  4  X  .2226  =  2.8938  tons. 

Let  the  coefficient  of  shearing  strength  be  4  tons  per  square 
inch,  and  suppose  that  the  surfaces  of  the  angle-irons  and  of 
the  flange  are  close  together ;  then 


2.8038 
area  of  rivet  =  — =    -  =  .7234  sq.  in., 

4 


and  its  diameter  =  .96  in. 

If  the  surfaces  are  not  close  together,  so  that  the  rivet  may 
be  subjected  to  a  bending  action,  then,  by  Ex.  3,  the  average 
intensity  of  shear  in  a  section  =1.4=3  tons  per  sq.  in.,  and 
hence 

area  of  rivet  =  — — -  =  .9646  sq.  in. ; 
its  diameter  is  i.i  in. 


THEORY   OF  STRUCTURES. 


13.  Beam  acted  upon  by  Forces  Oblique  to  its  Direc- 
tion, but  lying  in  a  Plane  of  Symmetry.  —  In  discussing  the 
equilibrium  of  such  a  beam  the  forces  may  be  resolved  into 
components  parallel  and  perpendicular  to  the  beam,  and  their 
respective  effects  superposed. 


FIG.  304. 

Let  AB  be  the  beam,  Pt ,  P2,  Pz ,  . .  .  the  forces,  and  al ,  <*a , 
aa  the-lr  respective  inclinations  to  the  neutral  axis. 

Divide"-the  beam  into  any  two  segments  by  an  imaginary 
plane  MN  ^perpendicular  to  the  beam,  and  consider  the  seg- 
ment A  MN. 

It  is  kept  in  equilibrium  by  the  external  forces  on  the  left 
of  MN  and  by  the  elastic  reaction  of  the  segment  BMN  upon 
the  segment  AMN  at  the  plane  MN. 

The  resultant  force  along  the  beam  is  the  algebraic  sum  of 
the  components  in  that  direction,  of  Pl ,  Pt ,  P3 ,  .  .  .  , 


=  Pl  cos  a1  +  P2  cos  «,-{-•••  =  -SX/*  cos  a). 

It  may  be  assumed  that  this  force  acts  along  the  neutral 
axis,  and  is  uniformly  distributed  over  the  section  MN. 

Thus,  if  A  is  the  area  of  the  section,  -— — - — ^  is  the  in- 
tensity of  stress  due  to  this  force. 

Again,  the  components  of  P1 ,  P% ,  P3 ,  . .  . ,  perpendicular  to 
the  beam,  are  equivalent  to  a  single  force  and  a  couple  at  MN. 

The  single  force  at  MN  is  the  Shearing  Force,  is  per- 
pendicular to  the  beam,  and  is  the  algebraic  sum  of  Pl  sin  al9 
Pz  sin  «2,  .  .  .  , 

=  Pl  sin  al  -\-  P^  sin  ora  +  .   .  .   =  2"(P  sin  a). 


BEAM  ACTED    UPON  BY  OBLIQUE   FORCES.  397 

This    force    develops    a    mean    tangential    unit    stress    of 

— 2 — - — —  in  MN,  and  deforms  the  beam,  but  so  slightly  as  to 
A 

be  of  little  account. 

The  moment  of  the  couple  is  the  algebraic  sum  of  the 
moments  with  respect  to  MN  of  Pl  sin  a, ,  P2  sin  oc^ ,  .  .  .  , 

=  Pl  sin  #,/!  +  PI  sin  a^p^  +  .  .  .  =  2(Pp  sin  a), 

/,,/„»•••   being  respectively  the  distances  of  the  points  of 
application  of  P, ,  P9 ,  .  .  .  from  MN. 

Now,  2(Pp  sin  at)  is  the  resultant  moment  of  all  the  external 
forces  on  the  left  of  MN,  for  the  resultant  moment  of  the  com- 
ponents along  the  beam  is  evidently  nil.  Hence, 


/„  -  *2(PJ>  sin  a), 

is  the  unit  stress  in  the  material  of  the  beam  at  a  distance  y 
from  the  neutral  axis  due  to  the  bending  action  at  MN  of  the 
external  forces  on  the  segment  AMN. 

Hence,  also,  the  total  um\.  stress  in  the  material  in  the  plane 
MN  at  a  distance  y  from  the  neutral  axis  is 

2(P  cos  a)  2(P  cos  a) 

±-~-   2±fy=±~   --   2± 


the  signs  depending  upon  the  kind  of  stress. 

It  will  be  observed  that  this  formula  is  composed  of  two 
intensities,  the  one  due  to  a  direct  pull  or  thrust,  the  other  due 
to  a  bending  action.  The  latter  is  proportional  to  the  distance 
of  the  unit  area  under  consideration  from  the  neutral  axis.  It 
is  sometimes  assumed  that  the  same  law  of  variation  of  stress 
holds  true  over  the  real  or  imaginary  joints  of  masonry  and 
brickwork  structures,  e.g.,  in  piers,  chimney-stacks,  walls, 
arches,  etc.  In  such  cases  the  loci  of  the  centres  of  pressure 
correspond  to  the  neutral  axis  of  a  beam,  and  the  maximum 


THEORY  OF  STRUCTURES. 

and  minimum  values  of  the  intensity  occur  at  the  edges  of  the 
joint. 

EXAMPLE  I.  A  horizontal  beam  of  length  /,  depth  d,  and 
sectional  area  A  is  supported  at  the  ends,  and  carries  a  weight 
W  at  its  middle  point.  It  is  also  subjected  to  the  action  of  a 
force  H  acting  in  the  direction  of  its  length. 

First.  Let  the  line  of  action  of  //  coincide  with  the  axis  of 
the  beam. 

The  intensity  of  the  stress  in  the  skin  at  the  centre 

=  ±£±7". 

But     c  oc  d,     and     7  =  Ak*  oc  Ad\ 


I       .  ,     Ad 
.-.  -  a  Ad—  — , 
c  n 


n  being  a  coefficient  depending  upon  the  form  of  the  section. 

If  the  section  is  a  circle,  n  =  8  ;  if  a  rectangle,  n  =  6. 
Hence, 

HI     ,   n  W  A 

the  skin  stress  =  db  ~2  I  *  4~  ~  ~ZF  3)» 
si  \        4  •"  d/ 

Wl 

since  M  = . 

4 

Wl 
If  the  load  W7  is  uniformly  distributed,  M  =  -5-. 

o 

Thus,  a  very  small  load  on  the  beam  may  considerably  in- 
crease the  intensity  of  stress,  and  this  intensity  will  be  still 
further  increased  by  the  deflection  of  the  beam  under  its  load, 
so  that,  in  order  to  prevent  excessive  straining,  it  is  often 
necessary  to  introduce  more  supports  than  are  actually  required 
to  make  the  beam  sufficiently  stiff. 

Second.  If  the  line  of  action  of  H  is  at  a  distance  h  from 
the  neutral  axis,  an  additional  bending  moment  Hh  will  be  in- 
troduced. 


BEAM  ACTED    UPON  BY  OBLIQUE  FORCES. 


399 


Ex.  2.  The  inclined  beam  OA,  carrying  a  uniformly  dis- 
tributed load  of  w  per  unit  of  length,  is  supported  at  A  and  rests 
against  a  smooth  vertical  surface  at  O. 

The  resultant  weight  wl  is  vertical 
and  acts  through  the  centre  C  of  OA  ; 
the  reaction  R^  at  O  is  horizontal. 

Let  the  directions  of  wl  and  R^  meet 
in  B.  For  equilibrium,  the  reaction  ^3 
at  A  must  also  pass  through  B. 

Let  the  vertical  through  C  meet  the 
horizontal  through  A  in  D. 

The  triangle  ABD  is  a  triangle  of  forces  for  the  three  forces 
which  meet  at  B. 


FlG<  3°5' 


R 

-  / 

wl 


AD        AD 


n  T\   —  T* 

BD      2.  DC 


Of, 


being  the  angle  OAD.     Hence 


wl 
R,  —  —  cot  a. 


Consider  a  section  MN,  perpendicular  to  the  beam,  at  a  dis- 
tance x  from  O. 

The  only  forces  on  the  left  of  MN  are  R1  and  the  weight 
upon  OM.  This  last  is  wx,  and  its  resultant  acts  at  the  centre 

x 
of  OM,  i.e.,  at  a  distance  -  from  MN. 

The  component  of  Rt  along  the  beam 

wl  cos2  a 

=  R^  cos  a  =  --  :  -  . 
2    sin  a 


The  component  of  R^  perpendicular  to  the  beam 


D    •  wl 

=  Rl  sm  a  =  —  cos  a. 


40O  THEORY   OF  STRUCTURES. 

The  component  of  wx  along  the  beam  =  wx  sin  a. 

The  component  of  wx  perpendicular  to  the  beam  =  wxcos  a. 

Hence, 

wl  cosa  a 

the  total  compression   at  NM  = : -4-  wx  sin  a  =  C- ; 

2    sin  or 


the  shearing  force  at  MN  =  —  cos  a  —  wx  cos  a  =  Sx 


the  bending  moment  at  MN=  x  —  cos  a  —  —  wx  cos  a  =  Mx\ 
and 


These  expressions  may  be  interpreted  graphically  as  already 
described,  Cx  ,  Sx  being  represented  by  the  ordinates  of  straight 
lines,  and  Mx ,  fy  by  the  ordinates  of  parabolas. 

fy ,  for  example,  consists  of  two  parts  which  may  be  treated 
independently.  Draw  OE  and  AF 
perpendicular  to  OA,a.nd  respectively 
equal  or  proportional  to 

wl  cos3  a  wl  cos2  a       wl 

— -7  —. and  — -  — : +  —r  sin  a. 

2A  sin   a  2A  sin  a    '    A 

Join  EF.     The  unit  stress  at  any 
FlG-  3°6-  point  of  the  beam  due  to  direct  com- 

pression is  represented  by  the  ordinate  (drawn  parallel  to  OE 
or  AF)  from  that  point  to  EF. 

Upon  the  line  GG'  drawn  through  the  middle  point  B  per- 
pendicular to  OA,  take  .BG  =  BG ',  equal  or  proportional  to 

y  wl* 

y-r-  cos  a.    According  as  the  stress  due  to  the  bending  action 

1      o 

at  any  point  of  the  beam  is  compressive  or  tensile,  it  is  repre- 
sented by  the  ordinate  (drawn  parallel  to  OE  and  AF)  from 
that  point  to  the  parabola  OGA  or  OG'A  ;  G  and  £',  respec- 
tively, being  the  vertices,  and  GG'  a  common  axis. 


SIMILAR   GIRDERS— PRINCIPAL   PROPERTIES.          4OI 

By  superposing  these  results,  the  parabolas  EHF,  EH'F are 
obtained,  the  ordinates  of  which  are  respectively  proportional 
to  the  values  of  fy  for  the  compressed  and  stretched  parts  of 
the  beam,  i.e.,  for  the  parts  above  and  below  the  neutral! 
surface. 

14.  Similar  Girders. — Two  girders  are  said  to  be  similar 
when  the  linear  dimensions  of  the  one  bear  the  same  constant 
proportion  to  the  corresponding  linear  dimensions  of  the  other. 

Thus,  if  ft,  ft',  d,  $',  A,  X'  are  corresponding  breadths,  depths, 
and  lengths  of  two  similar  girders, 

ft        6        X 

~8'  =  <T'  ~  A?  =  a  constant  "  /*»  suPpose. 

15.  To    Deduce    the    Principal    Properties   of  Similar 
Girders. — (a)  The  weight   of   a  girder  is  proportional  to  the 
product   of   an   area  and  length,  i.e.,  to  the  cube  of  a  linear 
dimension. 

Hence,  the  weights  of  similar  girders  vary  directly  as  the 
cubes  of  their  linear  dimensions.  Hence,  too,  the  unit  stresses 
must  vary  directly  as  their  linear  dimensions. 

(b)  The  Breaking  Weight  of  a  girder  is  calculated  from  a 

formula  of  the  form  W '=  S—y-j  a  being  an  area, da.  depth,  and 
/  a  length. 

Now  -  is  constant  for  similar  girders,  so  that  W  is  propor- 
tional to  a,  i.e.,  to  the  square  of  a  linear  dimension. 

Hence,  the  Breaking  Weights  of  similar  girders  vary  directly 
as  the  squares  of  their  linear  dimensions. 

EXAMPLE. — A  girder  resting  upon  two  supports  80  ft. 
apart  is  10  ft.  deep  and  weighs  6  tons.  Determine  the  length 
and  depth  of  a  similar  girder  weighing  48  tons. 

flengthV      /depth y      48 
\to~/==\IS/  ~-~~  6~r=8' 

Hence,  the  length  =  160  ft.  and  the  depth  =  20  ft.  Also, 
the  unit  stresses  are  in  the  ratio  of  10  to  20,  and  the  breaking 
weights  in  the  ratio  of  io2  to  2O2. 


4°2  THEORY  OF  STRUCTURES. 

16.  To  Discuss  the  Relations  between  the  Correspond- 
ing Sectional  Areas,  Moments  of  Inertia,  Weights,  Bend- 
ing Moments,  etc.,  of  two  Girders  which  have  the  same 
Sectional  Form  and  are  thus  related  : 

The  forces  upon  the  one  being  Pl  ,  P9  ,  P9  ,  .  .  .  with  abscissae 
x^  xt,  xtt  .*  .  .  those  upon  the  other  are  nPl  ,  nPt,  nP3,  .  .  . 
with  abscissae  pxl  ,  pxt  ,  px9  . 

The  spans  and  corresponding  lengths  are  in  the  constant 
ratio  /. 

Corresponding  sectional  breadths  are  in  the  constant  ratio  q. 
Corresponding  sectional  depths  are  in  the  constant  ratio  r. 
Let  A,  A'  be  corresponding  sectional  areas  ; 

/,  I'      "  "  moments  of  inertia  ; 

<2,  0',   "  "  weights  ; 

S,  Sf     "  "  shearing  forces  ; 

M,  M'  "  "  bending  moments  ; 

/,/'      "  "  flange  unit  stresses  ; 

j,  s'       "  "  web  unit  stresses  ; 

R,  R'    "  "  radii  of  curvature  ; 

A,  4f    "  "  deflections  ; 

W,  W  "  "  breaking  weights. 

/.  («)  A  a  product  of  a  breadth  and  depth  ; 


(/3)  I  a  product  of  a  breadth  and  the  cube  of  a  depth  ; 

.-./'=  Iqr\ 
(y)  Q  a  product  of  a  length,  breadth,  and  depth  ; 

...  Q'=  Qn=  Qpqrp, 

p  being  the  ratio  of  the  specific  weights  of  the  materials  of  the 
girders. 

If  the  materials  are  the  same, 

p  =  i     and     n  =  pqr. 


SIMILAR   GIRDERS.  4°3 

($}  S'  =  Sn  =  Spqrp,  for,  from  (y),  n  =  pqrp. 
(e)  M  is  the  product  of  a  force  and  a  length 
/.  M'  =  Mnp  —  Mp*qrp, 

cM  •        c'M' 

(C)  f=-f     and    /'  =  —-; 


f      c'M'  I  i        / 

.«.  -^—  =  —  —  -  —  =  rnp  —  ,  =  —p. 
f       c  M  I'  ^qr*       r  r 

(rf)  s  =  -      and     *'  =     7  ; 


E      f  E'      f 

(6)   —  =  -      and      ^7-  =  —  ,  E  and  £'   being  the   coef 

JK.  C  £\.  C 

ficients  of  elasticity  of  the  respective  girders  ; 


____    __        _   _. 

"  ~R  ~ff  c   E~~~~  E     p*p  ~=  E  /2  p9 

(a  length)2 

(0    ^  is  proportional  to  —  j:  -  7  -  -  -  ; 
v  J  radius  of  curvature 


the  product  of  an  area  and  depth 
(K)   Wis  proportional  to-  a  length  ~J 

W_   _  qr.r  _  q^_ 

w~-     p     '-  p' 

Hence,  the  values  of  A',  /',  Qf,  .  .  .  may  be  derived  from 
tho  >e  of  Aj  /,  Q,  .  .  .  by  means  of  certain  constant  multipliers. 


404  THEORY  OF   STRUCTURES. 

Cor.   I.  If  the   two   girders   are  similar  and   of   the  same 
material, 

p  =  g  =  r  =  ju,     E  =  J5',     and     p=i. 

Hence, 

from  (y),    Q  =  <2/A    and   the   weights   vary   directly  as   the 
cubes  of  the  linear  dimensions ; 

"  (e),  M'  =  MJJ?,  and  the  bending  moments  vary  directly 
as  the  fourth  powers  of  the  linear  di- 
mensions; 

/'  s/ 

"     (C)  and  (;?),  -j  —  P  =  — ,  and  the  flange  unit  stresses  vary 
/  s 

directly  as  the  web  unit  stresses; 


"     (z)>     "A  =  ^8>  and   t^ie  deflections  vary  directly  as  the 
squares  of  the  linear  dimensions  ; 

Wr 
"     W  ^M7  ==  ^ '  an<^  ^e  breaking  weights  vary  directly  as 

the  squares  of  the  linear  dimensions. 

Cor.  2.  Let  the  girders  be  of  the  same  material,  of  equal 
length,  of  equal  rectangular  sectional  areas,  and  equally  loaded. 

Let  b,  blt  and  d,  d,,  respectively,  be  the  breadths  and 
depths  of  the  girders.  Then 

bl  =  qb     and     dl  =  rd. 
Hence, 

b^d^  =  qrbd. 

But  b,d}  =  bd\  /    qr  —  i.     Also,  p  —  i. 


ALLOWANCE  FOR  THE   WEIGHT  OF  A  BEAM.  405 

f         i 

Thus  from  C,  y  =  -  =  q  ; 

R'  A 

«     0andi,     --  =  r'  =       ; 


«     ,-  ---  r  —  ~~ 

W  ~       ~  f  ' 

b 
If  dl  =  b,     then  bl  —  ^/,     and    r  =  -,. 


_ 

Hence,  --  -  ?>     and     ^  ==          =  = 

17.  To  make  Allowance  for  the  Weight  of  a  Beam.—  A 

beam  is  sometimes  of  such  length  that  its  weight  becomes  of 
importance  as  compared  with  the  load  it  has  to  carry,  and 
must  be  taken  into  account  in  determining  the  dimensions  of 
the  beam. 

The  necessary  provision  may  be  made  by  increasing  the 
width  of  the  beam  designed  to  carry  the  external  load  alone, 
the  width  being  a  dimension  of  the  first  order  in  the  expression 
for  the  elastic  moment. 

Assume  that  the  weight  of  the  beam  and  the  external  load 
are  reduced  to  equivalent  uniformly  distributed  loads. 

Let  Wt  be  the  external  load. 
"       be   "     "     breadth  of  a  beam  designed  to  support  this 

load  only.      } 

"      Be  "     "     weight  of  the  beam. 
"     W   "     "     total  load,  the  weight  of   the  beam  being 

taken  into  account. 

"       b     "     "     corresponding  breadth  of  the  beam. 
«     g     «     «  «  weight     "     "        " 

Then      W-B=We, 

b        B       W       W-  B  We 


and 


be~  B~  w,  "  W.-B.  ~  we-  B; 


4O6  THEORY  OF  STRUCTURES. 

wj>.  wjs.  w; 

Hence,  6,-.—-^-,     B    --  -^—^     and     W    --  W^J; 

EXAMPLE.  —  Apply  the  preceding  results  to  a  cast-iron 
girder  of  rectangular  section  resting  upon  two  supports  30  ft. 
apart.  The  girder  is  12  in.  deep  and  carries  a  uniformly  dis- 
tributed load  of  30,000  Ibs. 

Take  4  as  a  factor  of  safety  ;  be  is  given  by 

I2OOOO  bd* 


where  C  —  30,000  Ibs.,     d  =  12  in.,     and     /  =  360  in.  ; 

/.  be  —    5  in. 


Hence, 


Be  =  -^7  -^  X  30  X  450  =  5625  Ibs. ; 
144 

We  —  Be  —  30000  —  5625  =  24,375  Ibs. ; 


30000,5 
24375 


lbs> ; 

W=  ^      ^ 


EXAMPLES. 

1.  An  iron  bar  is  bent  into  the  arc  of  a  circle  of  500  ft.  diameter; 
the  coefficient  of  elasticity  is  30,000,000  Ibs.     Find  the  moment  of  resist- 
ance of  a  section  of  the  bar  and  the  maximum  intensity  of  stress  in  the 
metal,  (a)  when  the  bar  is  round  and  i  in.  in  diameter,  (<£)  when  the  bar 
is  square  having  a  side  of  r  inch. 

If  the  metal  is  not  to  be  strained  above  10,000  Ibs.  per  sq.  in.,  find  (c} 
the  diameter  of  the  smallest  circle  into  which  the  bar  can  be  bent. 

Ans.—(a)  ^Tt  in. -Ibs.;  5000 Ibs.  (b)  833^  in. -Ibs.;  5000  Ibs.    (c)  250  ft. 

2.  A  piece  of  timber  10  ft.  long,  12  in.  deep,  8  in.  wide,  and  having  a 
working  strength  of  1000  Ibs.  per  sq.  in.,  carries  a  load,  including  its 
own  weight,  of  w  Ibs.  per  lineal  foot.     Find  the  value  of  w,  (a)  when  the 
timber  acts  as  a  cantilever ;  (b)  when  it  acts  as  a  beam  supported  at  the 
ends.     Find  (c)  stress  in  material  3  in.  from  neutral  axis  at  fixed  end  of 
cantilever  and  at  middle  of  beam. 

Am, — (a)  320  Ibs.;  (b)  1280  Ibs.;  (c]  500  Ibs.  per  sq.  in. 

3.  Is  it  safe  for  a  man  weighing  160  Ibs.  to  stand  at  the  centre  of  a 
spruce  plank  loft,  long,  2  in.  wide,  and  2  in.  thick,  supported  by  vertical 
ropes  at  the  ends  ?     The  safe  working  strength  of  the  timber  is  1200  Ibs. 
per  sq.  in. 

Ans.  No  ;  the  maximum  safe  weight  at  the  centre  is  53^  Ibs. 

4.  Compare  the  uniformly  distributed  loads  which  can  be  borne  by 
two  beams  of  rectangular  section,  the  several  linear  dimensions  of  the 
one  being  n  times  the  corresponding   dimensions  of  the  other.     Also 
compare  the  moments  of  resistance  of  corresponding  sections. 

Ans.  n* ;  n*. 

5.  A  cast-iron  beam  of  rectangular  section,  12  in.  deep,  6  in.  wide,  and 
16  ft.  long,  carries,  in  addition  to  its  own  weight,  a  single  load  P\  the 
coefficient  of  working  strength  is  2000  Ibs.  per  sq.  in.     Find  the  value  of 
P  when  it  is  placed  (a)  at  the  middle  point;  (b)  at  2$-  ft.  from  one  end. 

Ans.— (a)  8475  Ibs. ;  (b)  11,300  Ibs. 

6.  A  round  and  a  square  beam  of  equal  length  and  equally  loaded  are 
to  be  of  equal  strength.  Find  the  ratio  of  the  diameter  to  the  side  of  the 
square.  Ans.   \/~$6  : 

407 


408  THEORY  OF  STRUCTURES. 

7.  Compare  the  relative  strengths  of  two  beams  of  the  same  length  and 
material  (a)  when  the  sections  are  similar  and  have  areas  in  the  ratio 
of  i  to  4;  (b}  when  one  section  is  a  circle  and  the  other  a  square,  aside  of 
the  latter  being  equal  to  the  diameter  of  the  former. 

Ans.  (a)  i  to  8 ;  (b)  56  to  33. 

8.  Compare  the  strength  of  a  cylindrical  beam  with  the  strength  of 
the  strongest  (a)  rectangular  and  (b)  square  beam  that  can  be  cut  from  it. 

Ans.  (a)  112  :  99  |/^~;  (b)  33  :  14  fa 

9.  A  boiler-plate  tube  36  ft.  long,  30  in.  inside  diameter,  weighs  4200 
IDS.  and  rests  upon  supports  33  ft.  apart.     Find  the  maximum  intensity 
of  stress  in  the  metal.     What  additional  weight  may  be  suspended  from 
the  centre,  assuming  that  the  stress  is  nowhere  to  exceed  8000  Ibs.  per 
sq.  in.  ? 

Ans.  741  i-  Ibs.  per  sq.  in.;  1 8,854  -|f|  Ibs. 

10.  Compare  the  relative  strengths  of  two  rectangular  beams  of  equal 
length,  the   breadth  (ft)  and   depth  (d)  of  one   being  the  depth  (b)  and 
breadth  (d)  of  the  other.  Ans.  d*  :  P. 

11.  A  yellow-pine  beam  14  in.  wide,  15  in. deep,  and  resting  upon  sup- 
ports 129  in.  apart,  was  just  able  to  bear  a  weight  of  34  tons  at  the  cen- 
tre.    What  weight  will  a  beam  of  the  same  material,  of  45  in.  span  and 
5  in.  square,  bear  ?  Ans.  \\\  tons. 

12.  A  cast-iron   rectangular  girder  rests  upon  supports  12   ft.  apart 
and  carries  a  weight  of  2000  Ibs.  at  the  centre.     If  the  breadth  is  one- 
half the  depth,  find  the  sectional  area  of  the  girder  so  that  the  intensity 
of  stress  may  nowhere  exceed  4000  Ibs.  per  sq.  in. 

Ans.   1 8  sq.  in.;  if  weight  of  girder  is  to  be  taken  into  account, 
the  depth  d  is  given  by  ds  —  i.oi2$d*  —  216  =  o. 

13.  Find  the  depth  of  a  wrought-iron  girder  6  in.  wide  which  might  be 
substituted  for  the  cast-iron  girder  in  the  preceding  question,  the  coeffi- 
cient of  strength  for  the  wrought-iron  being  8000  Ibs.  per  sq.  in. 

Ans.  4.762  in.;  if  weight  of  girder  is  to  be  included, the  depth  d 
is  given  by  d*  —  .54</2  —  108  =  o. 

14.  An  oak  beam  of  circular  section  and  22  ft.  long  is  strained  to  the 
elastic  limit  (2  tons  per  sq.  in.)  by  a  uniformly  distributed  load  of  2^ 
tons.     Find  the  diameter  of  the  beam.     What  load  2  ft.  from  one  end 
would  strain  the  material  to  same  limit?  Ans.  7  in.;  3T^V  tons. 

1 5.  A  uniform  beam  of  weight  W\  crossing  a  given  span  can  bear  a  uni- 
formly distributed  load  Wt..     What  load  may  be  placed  upon  the  same 
beam  if  it  crosses  the  span  in  n  equal  lengths  supported  at  the  joints  by 
piers  whose  widths  may  be  disregarded  ?        Ans.  n*(  W^  +  IVz)  —  W\. 


EXAMPLES.  409 

16.  A  flat  spiral  spring  .2  in.  wide  and  .03  in.  thick  is  subjected  to  a 
bending   moment  of  10  in. -Ibs.     Find  its  radius  of  curvature,  E  being 
56,000,000  Ibs.  Ans.  1.62  in. 

17.  Determine  the  diameter  of  a  solid  round  wrought-iron  beam  rest- 
ing upon  supports  60  in.  apart  and  about  to  give  way  under  a  load  of  30 
tons  at  14  in.  from  one  end.     Take  5  as  a  factor  of  safety  and  8960  Ibs. 
per  sq.  in.  as  the  safe  working  intensity  of  stress. 

Ans.  5.47  in.;  if  weight  of  beam  is  taken  into  account,  the  diam- 
eter (d)  is  given  by  2019584—  i375^2—  12320^"  =o. 

1 8.  A  wrought-iron  bar  i-|  in.  wide  and  20  ft.  long  is  fixed  at  one  end 
and  carries  a  load  of  500  Ibs.  at  the  free  end.     Find  the  depth  of  the  bar, 
so  that  the  stress  may  nowhere  exceed  10,000  Ibs.  per  sq.  in. 

Ans.  6.928  in.;  if  weight  of  bar  is  included,  the  depth  d  is  given 

by  d*  —  4.8^  —  48=0. 

19.  Compare  the  moments  of  resistance  to  bending  of  a  rectangular 
section  and  of  the  rhomboidal  and  isosceles  sections  which  can  be  in- 
scribed in  the  rectangle,  the  base  of  the  triangle  being  the  lower  edge 
of  the  rectangle.  Ans.  6  :  i  :  i  or  6  :  i  :  2. 

20.  A  stress  of  i  Ib.  per  sq.  in.  produces  a  strain  of  7^7777  in  a  beam 
12  in.  square  and  20  ft.  between  supports.     Find  the  radius  of  curvature 
and  the  central  deflection  under  a  load  of  2000  Ibs.  at  the  middle  point. 

Ans.   2400  ft. ;  \  in. 

21.  A  piece  of  greenheart  139  in.  between  supports,  9  in.  wide  and  8 
in.  deep,  was  successively  subjected  to  loads  of  4,  8,  and   16  tons  at  the 
centre,  the  corresponding  deflections  being  .32  in.,  .64  in.  and  1.28  in. 
Find  E  and  the  total  work  done  in  bending  the  beam. 

What  were  the  corresponding  inch-stresses  at  f  of  the  depth  of  the 
beam?          Ans.  E  =  5582682^  ;  1344  inch-tons  ;  ff  Ib.,  |f  Ib.,  Vr°  lb- 

22.  The  effective  length  of  the  Conway  tubular  bridge  is  412  ft.  ;  the 
effective  depths  of  a  tube  at  the  centre  and  quarter  spans  are  23.7  ft.  and 
22.25  ft-»  respectively  ;  the  sectional  areas  of  the  top  and  bottom  flanges 
are  respectively  645  sq.  in.  and  536  sq.  in.  at  the  centre  and  566   sq.  in. 
and  461  sq.  in.  at  the  quarter  spans  ;  the  corresponding   sectional  areas 
of  the  web  are  257  sq.  in.  and  241  sq.  in.     Assume  the  total  load  upon  a 
tube  to  be  equivalent  to  3  tons  per  lineal  foot,  and  that  the  continuity  of 
the  web  compensates  for  the  weakening  of  the  tension   flange  by  the 
rivet-holes.     Find  the  flange  stresses  and  the  deflection  at  the  centre  and 
quarter  spans,  E  being  24,000,000  Ibs.     What  will  be  the  increase  in  the 


4IO  THEORY  OF  STRUCTURES. 

central  flange  stresses  under  a  uniformly  distributed  live  load  of  £  ton 
per  lineal  foot  ? 

Ans.  At  centre  — —  =  181485  ;  ft  =  4.3799  tons  per  sq.  in.  ; 


Deflection  =  8.33  in. 

At  quarter  span  — —  =  132774  ;  ft  =  1.414       "       "        " 

fc  =1.256        "       "        " 

Deflection  =  6.25  in. 

The  stresses  and  deflections  are  increased  by  the  live  load  in 
the  ratio  of  5  to  4. 

23.  A  plate  girder  of  64  ft.  span  and  8  ft.  deep  carries  a  dead  load  of  2 
tons  per  lineal  foot.    At  any  section  the  two  flanges  are  of  equal  area,  and 
their  joint  area  is  equal  to  that  of  the  web.     Find  the  sectional  area  at 
the  centre  of  the  girder,  so  that  the  intensity  of  stress  in  the  metal  may 
not  exceed  3  tons  per  sq.  in.     The  deflection  of  the  girder  is  f  in.  at 
the  centre.     Find  E  and  the  radius  of  curvature. 

Ans.  128  sq.  in.  ;  15,360  ft.;  25,804,800  Ibs. 

24.  Taking  the  coefficient  of  direct  elasticity  at  15,000  tons,  the  coeffi- 
cient of  lateral  elasticity  at  60,000  tons,  and  the  limit  of  elasticity  at  10 
tons,  determine  the  greatest  deviation  from  the  straight  line  of  a  wrought- 

iron  girder  of  breadth  b  and  depth  d.  b* 

Ans.  —    — 7. 
24000^ 

25.  Find  the  stress  at  the  skin  and  also  at  a  point  4  in.  from  the  neu- 
tral axis  in  a  piece  of  10"  x8"  oak,  (a)  with  the  10"  side  vertical;  (b}  with 
the  8"  side  vertical.    The  oak  rests  upon  supports  3  ft.  apart  and  carries  a 
load  of  4900  Ibs.  at  its  middle  point.  Also  compare  (c)  the  strength  of  the 
beam  with  its  strength  when  a  diagonal  is  horizontal. 

Ans. — (a)  330! ;  I32T37  Ibs.  per  sq.  in. 


_ 
(c)  4  :  1/41  or  5  :  4/41. 

26.  Find  the  uniformly  distributed  load   which  can   be  borne  by  a 
rolled  T-iron  beam,  6"  X4"  x  £",  10  ft.  long,  fixed  at  one  end  and  free  at 
the  other,  the  coefficient  of  strength  being  10,000  Ibs.  per  sq.  in. 

Ans.  438  Ibs. 

27.  One  of  the  tubes  of  the  Britannia  bridge  has  an  effective  length  of 
470  ft. ,  depth  of  27^  ft. ,  and  deflects  1 2  in.  at  the  centre  under  a  uniformly 
distributed  load  of  1587  tons.     Find  E  and  the  central  flange  stresses,  the 
sectional  areas  of  the  top  flange,  bottom  flange,  and  web  being  648  sq.  in., 
585  sq.  in.,  and  302  sq.  in.,  respectively. 

Ans.  E  ==  22,910,496  Ibs.;  ft  =  5.37  tons  per.  sq.  in.; 


EXAMPLES.  411 

28.  Find  the  moment  of  resistance  to  bending,  of  a  steel  I-beam,  each 
flange  consisting  of  a  pair  of  3-in.  x  3  in.  x  ^  in.  angle-irons,  riveted   to 
a  12  in.  x  f  in.  web,  the  coefficient  of  strength  being  5  tons  per  sq.  in. 
What  load  will  the  beam  carry  at  5  ft.  from  one  end,  its  span  being   20 
ft.  ?     Find  the  central  deflection,  and  also  the  deflection  at  the  loaded 
point,  E  being  15,000  tons. 

Ans.  287!^  in. -tons  ;  6f£f-  tons  disregarding  weight  of  beam,  or 
SyVA  tons  if  weight  of  beam  is  taken  into  account ;  deflection 
at  centre  =  f  in.,  at  loaded  point  =  ^  in. 

29.  A  shaft  5i  in.  deep  x  5  in.  wide  x  98  in.  long  has  one  end  abso- 
lutely fixed,  while  at  the  other  a  wheel  turns  at  the  rate  of  270  revolutions 
per  minute  ;  a  weight  of  200  Ibs.  is  concentrated  in  the  rim,  its  C.  of  G. 
being  2^  ft.  from  the  axis  of  the  shaft.     Find  the  maximum  stress  in  the 
material  of  the  shaft,  and  also  find  the  maximum  deviation  of  the  shaft 
from  the  straight,  E  being  27,000,000  Ibs. 

30.  The  square  of  the  radius  of  gyration  of  the  equal-flanged  section 
of  a  wrought-iron   girder  of  depth  d  is  rV^2;   the  area  of  the  section 
=  |d?2 ;  the  span  =  50  ft.     In  addition  to  its  own  weight  it  carries  a  uni- 
formly distributed  load  of  i¥V  Ibs.  per  lineal  foot ;  the  maximum  intensity 
of  stress  =  10,000  Ibs.  per  sq.  in.     Find  the  depth.     Also  determine  the 
stiffness,  E  being  25,000,000  Ibs.  Ans.   3!  in.;  Tfy. 

31.  The  central  section  of  a  cast-iron  girder  is  io£  in.  deep;  its  web 
area  is  five  times  the  area  of  the  top  flange,  and  the  moment  of  resistance 
of  the  section  is  360,000  in.-lbs.;  the  tensile  and  compressive  intensities  of 
stress  are  3000  and  7500  Ibs.  per  sq.  in.,  respectively.     Find  the  span  and 
load  so  that  the  girder  may  have  a  stiffness  =  .001,  E  being  17,000,000 
Ibs. 

Ans.  <zi  =  12^-  sq.  in.;  a*  =  i|£|  sq.  in.;  as  +  at  =  97|||  sq. 
in.;  span  =  136  ft.;  uniformly  distributed  load  =  1764^  Ibs. 

32.  A  double-flanged  cast-iron  girder  14  in.  deep  and  20  ft.  between  sup- 
ports carries  a  uniformly  distributed  load  of  20  tons.     Find  suitable  di- 
mensions for  the  section,  the  tensile  and  compressive  inch-stresses  being 
2  tons  and  5  tons,  respectively.     Also  find  the  stiffness  of  the  beam,  E 
being  8000  tons. 

Ans.  Let  thickness  of  web  =  i  in.;  a\  =  22f|  sq.  in.;  a*  =  4-^.- 
sq.  in.;  stiffness  =  .001875. 

33.  The  deflection  of  a  uniformly  loaded  horizontal  beam  supported  at 
the  ends  is  not  to  exceed  i  in.  in  50  feet  of  span,  and  the  stress  in  the  ma- 
terial is  not  to  exceed  400  Ibs.  per  sq.  in.     Find  the  ratio  of  span  to 
depth,  E  being  1,200,000  Ibs.  per  sq.  in.,  and  the  neutral  axis  being  at 
half  the  depth  of  the  beam.  Ans.  20. 

34.  Two  equal  weights  are  placed  symmetrically  at  the  points  of  tri- 
section  of  a  beam  of  uniform  section  supported  at  the  ends.   These  weights 


412  THEORY  OF  STRUCTURES. 

are  then  removed  and  other  two  equal  weights  are  placed  at  the  quarter 
spans.  Find  the  ratio  of  the  two  sets  of  weights  so  that  the  maximum 
intensity  of  stress  may  be  the  same  in  each  case.  Also  show  that  the 
stiffness  of  the  beam  is  the  same  in  each  case.  Ans.  3  to  4. 

35.  A  cast-iron  beam  has  a  cruciform  section  with  equal  ribs  2  in. 
thick  and  4  in.  long.     If  the  intensity  of  longitudinal  shear  at  the  neutral 
axis  is  i  ton  per  sq.    in.,  find    the  total    shear  which  the  section  can 
bear,  and  also   find  the  moment  of   resistance,  the  least   coefficient  of 
working  tensile  and  compressive  stress  being  i  ton  per  sq.  in. 

Ans.   59.31  tons;  34.6  tons. 

36.  If  a  spiral  spring  is  fastened  to  the  barrel  so  that  there  is  no 
change  of  direction  relatively  to  the  barrel,  show  that  the  tendency  to 
unwind  is  directly  proportional  to  the   amount  of  winding  up.     (Con- 
dition of  perfect  isochronism.) 

37.  Show  that  the  modulus  of  rupture  vi  any  material  is  18  times  the 
load  which  will  break  a  beam  12  in.  long,  i  in.  deep,  and  i  in.  wide  when 
applied  at  the  centre. 

38.  Find  the  limiting  length  of  a  wrought-iron  cylindrical  beam  4  in. 
in  diameter,  the  modulus  of  rupture  being  42,000  Ibs.      What  uniformly 
distributed  load  will  break  a  cylindrical  beam  of  the  same  material  20  ft. 
long  and  4  in.  in  diameter  ?  Ans.  64.8  ft.  ;  8800  Ibs. 

39.  A  red-pine  beam  18  ft.  long  has  to  support  a  weight  of  10,000  Ibs. 
at  the  centre.     The  section  is  rectangular  and  the   depth   is  twice  the 
breadth.     Find  the  transverse  dimensions,  the  modulus  of  rupture  being 
3500  Ibs.,  and  10  being  a  factor  of  safety.     (Neglect  the   weight  of   the 
beam.)  Ans.  b  =  9.84  in.  ;  d=  19.68  in. 

40.  A  round  oak  cantilever  10  ft.  long  is  just  broken  by  a  load  of  600 
Ibs.  suspended  from  the  free  end.    Find  its  diameter,  the  modulus  of  rup- 
ture being  10,000  Ibs,     (Neglect  the  weight  of  the  beam.) 

Ans.  4.185  in. 

41.  Determine  the  breaking  weight  at  the  centre  of  a  cast-iron  beam 
of  6  ft.  span  and  4  in.  square,  the  coefficient  of  rupture  being  30,000  Ibs. 

Ans.  26,666|  Ibs. 

42.  The  flooring  of  a  corn  warehouse  is  supported  upon  yellow-pine 
joists  20  ft.  in  the  clear,  8  in.  wide,  10  in.  deep,  and  spaced   3  ft.  centre 
to  centre.     Find  the    height  to  which  corn  weighing  48^  Ibs.  per  cu. 
ft.  may  be  heaped   upon  the  floor,  10  being  a  factor  of  safety  and  3000 
Ibs.  the  coefficient  of  rupture.  Ans.   .68  ft. 

43.  A  yellow-pine  beam  14  in.  wide,  15  in.  deep,  and  resting  upon  sup- 
ports  126  in.  apart,  broke  down  under  a  uniformly  distributed  load  of 
60.97  tons.     Find  the  coefficient  of  rupture.  Ans.  2731.456. 


EXAMPLES.  413 

44.  Find  the  breaking  weight  at  the  centre  of  a  Canadian  ash  beam 
2-J-  in.  wide,  3!  in.  deep,  and  of  45  in.  span,  the  coefficient  of  rupture  being 
7250.  Ans.  4934uV 

45.  A  timber  beam  6  in.  deep,  3  in.  wide,  96  in.  between  supports,  and 
weighing  50  Ibs.  per  cu.  ft.,  broke   down   under  a  weight  of   10,000  Ibs. 
at  the  centre.     Find  the  coefficient  of  rupture.  Ans.  891  1^. 

46.  A  wrought-iron  bar  2  in.  wide,  4  in.  deep,  and  144  in.  between  sup- 
ports, carries  a  uniformly  distributed  load   W  in  addition  to  its   own 
weight.     Find  W,  4  being  a  factor  of  safety  and   50,000  Ibs.  the  coeffi- 
cient of  rupture.  Ans.  5235!  Ibs. 

47.  Find   the   length  of  a  beam  of  Canadian  ash  6  in.  square  which 
would  break  under  its  own  weight  when  supported  at  the  ends.     The 
coefficient  of  rupture  =  7000  Ibs.,  and  the  weight  of  the  timber  =  30  Ibs. 
per  cu.  ft.  Ans.  230  ft. 

48.  The  teeth  of  a  cast-iron  wheel  are  3^  in.  long,  2^  in.  deep,  and  7 
in.  wide.     What  is  the  breaking  weight:  of  a  tooth,  the  coefficient  of 
rupture  being  5000  Ibs.  ?  Ans.  50,625  Ibs. 

49.  A  wrought-iron  bar  4  in.  deep,  £  in.  wide,  and  rigidly  fixed  at  one 
end  gave  way  at  32  in.  from  the  load  when  loaded  with  1568  Ibs.  at  the 
free  end.     Find  the  coefficient  of  rupture.  Ans.  4181^. 

50.  A  cast-iron  beam  12  in.  wide  rests  upon  supports  1  8  ft.  apart,  and 
carries  a  12-in.  brick  wall  which  is  12^  ft.  in  height  and  weighs  112  Ibs. 
per  cu.  ft.     Taking  63,000  as  the  modulus  of  rupture  for  a  uniformly 
distributed   load  and  5  as  a  factor  of  safety,  find  the  depth  of  the  beam, 
(a)  neglecting  its  weight  ;  (b)  taking  its  weight  into  account. 

Also  (c)  determine  the  depth  of  a  cedar  beam  which  might  be  sub- 
stituted for  the  cast-iron  beam,  taking  11,200  Ibs.  as  the  modulus  of 
rupture  for  the  cedar.  Ans.  (a)  6  in.  ;  (fr)  6|  in.  ;  (c)  14.23  in. 

51.  A  cast-iron  girder  27^  in.  deep,  rests  upon  supports  26  ft.  apart. 
Its  bottom  flange  has  an  area  of  48  sq.  in.  and  is  3  in.  thick.     Find  the 
breaking  weight  at  the  centre,  the  ultimate  tensile  strength  of  the  iron 
being  15,000  Ibs.  per  sq.  in.     (Neglect  the  effect  of  the  web.) 

Ans.  253,846^  Ibs. 

52.  A  beam  of  rectangular  section,  of  breadth  b  and  depth  d,  is  acted 
upon  by  a  couple  in  a  plane  inclined  at  45°  to  the  axis  of  the  section. 
Compare  the  moment  of  resistance  to  bending  with  that  about  either 
axis. 


' 


53.  A  2-in.  wrought-iron  bar  10  ft.  long  is  held  at  the  ends  and  is 
whirled  about  a  parallel  axis  at  the  rate  of  50  revolutions  per  minute. 
If  the  distance  between  the  axis  of  the  bar  and  the  axis  of  rotation  is 
10  ft.,  find  the  maximum  stress  to  which  the  material  is  subjected. 

Ans.    17148.5  Ibs.  per  sq.  in. 


4H     .  THEORY  OF  STRUCTURES. 

54.  A  block  of  ice  3  in.  wide  and  4  in.  deep  has  its  ends  resting  upon 
supports  30  in.  apart  and  carries  a  uniformly  distributed  load  of  4800 
Ibs.     An  increase  of  pressure  to  the  extent  of  1125  Ibs.  per  sq.  in.  lowers 
the  freezing  point  i°  F.     Assuming  that  the  ordinary  theory  of  flexure 
holds  good,  find  the  temperature  of  the  ice.  Ans.  30°  F. 

55.  Find  the  limiting  length  of  a  cantilever  of    uniform  transverse 
section,/  being  the  coefficient  of  strength,  k  the  ratio  of  length  to  depth, 
and  u>  the  specific  weight  of  the  material. 

Ans.  — ^— - >  n  being  a  coefficient  depending  upon  the  form  of 
the  section. 

56.  If  the  beam  in  the  preceding  question  is  to  be  supported  at  its  two 
ends,  what  will  its  limiting  length  be  ?  Ans.  —    ,     . 

57.  Find  the  limiting  length,of  a  cedar  cantilever  of  rectangular  sec- 
tion, k  being  40,  w  =  36  Ibs.  per  cu.  ft.,  and/  =  1800  Ibs.  per  sq.  in. 

Ans.  60  ft. 

58.  A  steel  cantilever  2  in.  square  has  an  elastic  strength  of  15  tons 
per  sq.  in.     What  must  its  limiting  length  be  so  that  there  may  be  no 
set  f  Ans.  23.4  ft. 

59.  Find  the  limiting  length  of  a  wrought-iron  beam  of  circular  sec- 
tion, k  being  64  and  the  elastic  strength  8  tons  per  sq.  in.    What  will  this 
length  be  if  a  beam  of  I-section,  having  equal  flange  areas  and  a  web 
area  equal  to  the  joint  area  of  the  flanges,  is  substituted  for  the  circular 
section  ?  Ans.  84  ft. ;  224  ft. 

60.  A  rectangular  cast-iron   beam    having   its    length,   depth,   and 
breadth  in  the  ratio  of  60  to  4  to  i,  rests  upon  supports  at  the  two  ends. 
Find  the  dimensions  of  the  beam  so  that  the  intensity  of  stress  under  its 
own  weight  may  nowhere  exceed  4500  Ibs.  per  sq.  in. 

Ans.  I  =  128  ft. ;  d  =  8-jV  ft. ;  b  =  2T^  ft. 

61.  A  beam  supported  at  the  ends  can  just  bear  its  own  weight  W\.o- 

W 
gether  with  a  single  weight  —  at  the  centre.  What  load  may  be  placed 

at  the  centre  of  a  beam  whose  transverse  section  is  similar  but  m*  as 
great,  its  length  being  n  times  as  great  ?  If  the  beam  could  support  only 
its  own  weight,  what  would  be  the  relation  between  m  and  n  ? 


(m*       m^  n 

Ans.    W ;    m  =  -. 

\n        2  )  2 


62.  The  flanges  of  a  rolled  joist  are  each  4  in.  wide  by  \  in.  thick ; 
the  web  is  8  in.  deep  by£  inch  thick.  Find  the  position  of  the  neutral 
axis,  the  maximum  intensities  of  stress  per  square  inch  being  10,000 
Ibs.  in  tension  and  8000 Ibs.  in  compression.  Ans.  k\.  =  3$ ;  hi  ~  4$. 


EXAMPLES.  415 

63.  A  continuous  lattice-girder  is  supported  at  four  points,  each  of  the 
side  spans  being  140  ft.  n  in.  in  length,  22  ft.  3  in.  in  depth,  and  weigh- 
ing .68  ton  per  lineal  foot.     On  one  occasion  an  excessive  load  lifted  the 
end  of  one  of  the  side  spans  off  the  abutment.     Find  the  consequent  in- 
tensity of  stress  in  the  bottom  flange  at  the  pier,  where  its    sectional 
area  is  127  sq.  in.  Ans.  2.3893  tons  per  sq.  in. 

64.  A  railway  girder  is  101.2  ft.  long,  22.25  ft-  deep,  and  weighs  3764 
Ibs.  per  lineal  foot.   Find  the  maximum  shearing  force  and  flange  stresses 
at  25  ft.  from  one  end  when  a  live  load  of  2500  Ibs.  per  lineal  foot  crosses 
the  girder. 

65.  A  floor  with  superimposed  load  weighs  160  Ibs.  per  sq.  ft.  and  is 
carried  by  tubular  girders   17  ft.   c.  to  c.  and  42  ft.  between  bearings. 
Find  the  depth  of  the  girders  (neglecting  effect  of  web),  the  safe  inch- 
stress  in  the  metal  being  9000  Ibs.,  and  the  sectional  area  of  the  tension 
flange  at  the  centre  32  sq.  in.  Ans.  24.99  in- 

66.  Design  a  timber  cantilever  of  approximately  uniform  strength  from 
the  following  data:  length  =  12  ft.;   square  section;  load  at  free  end 
=  i  ton  ;    coefficient  of   working   strength  =  |  ton  per  sq.  in.      What 
must  be  the  dimensions  at  the  fixed  and  free  ends  so  that  the  cantilever 
might  carry  an  additional  uniformly  distributed  load  of  2  tons  ? 

Ans.  Side  =  15.1    in.  at    fixed    end  and  =  10   in.  at  free  end; 
side  =  19.1    in.  at  fixed  end  and  =  £(19.1    in.)   at   free  end. 

67.  Show  that  the  curved  profile  of  a  cantilever  of  uniform  strength 
designed  to  carry  a  load  W  at  the  free  end,  is   theoretically   a  cubical 
parabola.     Also   show  that  by  taking  the  tangents  to  the  profile  at  the 
fixed  end  as  the  boundaries  of  the  cantilever,  a  cantilever  of  approxi- 
mately uniform   strength  is  obtained  having  a   depth  at  the  free  end 
equal  to  two-thirds  of  the  depth  at  the  fixed  6nd. 

68.  Design  a  wheel-spoke  33  in.  in  length  to  be  of  approximately  uni- 
form strength,  the  intensity  of  stress  being  4000  Ibs.  per  sq.  in.  ;  the  load 
at  the  end  of  the  spoke  is  a  force  of  1000  Ibs.  applied  tangentially  to  the 
wheel's  periphery,  and  the  section  of  the  spoke  is  to  be  (a)  circular,  (b) 
elliptical,  the  ratio  of  the  depth  to  the  breadth  being  2|. 

Ans.  —  (a)     Depth  at  hub  =  6.982    in.,  at  periphery  =  4.634  in. 
(b)          "       ••     «•     =  9.435     "     "  =  9-35  in. 

Breadth  "     "     =  3.7/4     "     "  =  3-74  in. 

69.  A  beam  of  17  ft.  span   is  loaded  with  7,  7,  11,  and   n   tons  at 
points  1,6,  ii,  and  15  ft.  from  one  end.     Determine  the  depths  at  these 
points,  the  beam   being  of  uniform   breadth  and  of  approximately  uni- 
form strength  ;  the  coefficient  of  working  strength  =  2  tons  per  sq.  in.; 
the  depth  of  the  section  of  maximum  resistance  to  bending  =  16  in. 

16065  277  x   i6s  1087  x  i6a 


Ans'   *  = 


i785 

670  x  i62 
and   d<?  = 


41 6  THEORY  OF  STRUCTURES. 

70.  Design  a  cantilever  10  ft.  long,  of  approximately  uniform  strength, 
to  carry  a  load  of  4000  Ibs.  at  the  free  end,  the  coefficient  of  strength 
being  2000  Ibs.  per  sq.  in.,  and  the  section  (a)  a  rectangle  of  constant 
breadth  and  12  in.  deep  at  the  fixed  end  ;  (b)  a  square. 

How  will  the  results  be  modified  if  it  is  to  carry  an  additional  uni- 
formly distributed  load  of  4800  Ibs.  ? 

Ans. — First,   (a]  b  —  10  in.,    d  at    free    end  =  6    in.  ;    (b)   side 

=  ty  1440  at  fixed  end  and  =   tyiSo  at  free  end. 
Second,  (a)  b  =  16  in.,  */at  free  end  =  6  in. ;  (b)  side  =  ty  2304 
at  fixed  end  and  =    ^288  at  free  end. 

71.  Design  a  cantilever  10  ft.  long,  of  constant  breadth,  and  of  ap- 
proximately uniform  strength  to  carry  a  uniformly  distributed  load  of  5000 
Ibs.  on  the  half  of  the  length  next  the  free  end,  the  intensity  of  stress 
being  2000  Ibs.  per  sq.  in.,  and  the  section  a  rectangle  12  in.  deep  at  the 
fixed  end.     What  must  the  dimensions  be  if  loco  Ibs.  are  concentrated 
at  30  in.  from  fixed  end  ? 

Ans.  b  =  9!  in. ;  d  at  centre  =  6.928  in. ;  at  free  end  =  o. 

b=  10  in. ;  depth  =8.66  in.  at  7^  ft.  from  free  end,  =6.928 
in.  at  centre,  and  =  o  at  free  end. 

72.  A  gallery  30  ft.  long  and  10  ft.  wide  is  supported  by  four  9  in.  by  5 
in.  cantilevers  spaced  so  as  to  bear  equal  portions  of  the  superincumbent 
weight.  What  load  per  square  foot  will  the  gallery  bear,  the  coefficient  of 
working  strength  being  700  Ibs.  per  sq.  in.  ?     Find  the  depth  of  cast-iron 
cantilevers  3  in.  wide  which  may  be  substituted  for  the  above,  the  coef- 
ficient of  working  strength  being  2000  Ibs.  per  sq.  in.     How  should   the 
•depth  vary  if  the  cantilevers  are  to  be  of  uniform  strength  ? 

Ans.  42  Ibs.;  d*  =,18.9;  variation  of  depth  for  cast-iron  canti- 
lever is  given  by  loood*  =  189^,  x  being  distance  from  free 
end. 

73.  A  span  of  60  ft.  is  crossed  by  abeam  hinged  at  the  points  of  trisec- 
tinn  and  fixed  at  the  ends  ;  the  beam  has  a  constant  breadth  of  3  in.  and 
is  to  be  of  uniform  strength  ;  the  intensity  of  stress  is  3  tons  per  sq.  in. 
Determine  the  dimensions  of  the  beam  when  a  load  of  ^  ton  Per  lineal 
loot  covers  (a)  the  whole  span  ;  (b)  the  centre  span. 

Ans. — (a)  Depth  at  support  =  20  in.,  at  centre  =  |/2a 
(d)       "  "       =  VSo  in.,  "        "      =  |/2oT 

74.  In  the  following  examples  determine  the  position  of  the  neutral 
axis,  the  moment  of  resistance  to  bending,  the  resistance  to  shear,  and 
the  ratio  of  the  maximum  to  the  average  intensity  of  shear,  the  co- 
efficients of  strength  being  4^  tons  per  sq.  in.  for  tension  and  compression 
and  3^  tons  per  sq.  in.  for  shear. 

(I)  A  rectangle  2  in.  wide  and  6  in.  deep. 

Ans.   At  centre  ;   54  in. -tons  ;  28  tons  ;  3  to  2. 


EXAMPLES. 


417 


(II)  A  circular  section  4  in.  in  diameter. 

Ans.  At  centre  ;   28.2  in.  -tons;  33  tons;  4  to  3. 

(III)  A  regular  hexagonal  section  with  a  diameter  (a)  vertical,  (b) 
horizontal,  a  being  a  side  of  the  hexagon. 


Ans.—  (a)  At  centre; 


(b)  At  centre  ;*9; 
16 


1888 


l  7  to  5. 


:  1.258. 


(IV)  A  triangular  section  6  in.  deep,  with  a  base  6  in.  wide,  the  sides 
being  equal.          Ans.  4  in.  from  vertex  ;  40.5  in.  -tons  ;  42^  tons  ;  3  to  2. 

(V)  A  double-tee  section  composed  of  a  3o-in.  x  f-in.  web  and  four 
angle-irons  each  5  in.  x  3^  in.  x  £  in. 

Ans.  At  centre  ;  1501.06  in.  -tons  ;  22.  36  tons:  4.916  to  I. 

(VI)  A  section  having  a  semicircular  top  flange   of  8  in.  external 
diameter  and  i  in.  thick,  a  web  14  in.  deep  and  i  in.  thick,  and  a  bottom 
flange  8  in.  wide  and  i  inch  thick. 

(VII)  A  section    having  a  semi-elliptic  top  flange  2  in.  thick,  the  in- 
ternal major  and  minor  axes  being  8  in.  (horizontally)  and  4  in.  ^ver- 
tically), respectively,  a  bottom  flange  8  in.  wide  and  2   in.  thick,  and  a 
web  10  in.  deep  and  2  in.  thick. 

(VIII)  A  section  having  a  semi-elliptic  top  flange  2  in.  thick,  the 
external  major  and    minor  axes  being  10  in.    (horizontally)  and   6  in. 
(vertically),  respectively,  a  trapezoidal  web  8  in.  deep  having  a  width  of  3 
in.  at  the  top  and  6  in.  at  the  bottom,  and  a  bottom  semicircular  flange 
of  10  in.  external  diameter  and  2  in.  thick. 

(IX)  The  sections  shown  by  Figs.  307,  308,  and  309. 


FIG.  309. 


Also  find  the  diameters  of  the  rivets/?  in  Fig.  i,  neglecting  the  weak- 
ening effect  of  the  rivet-holes  in  the  bottom  flange.  What  is  the  ratio 
of  the  maximum  tensile  and  compressive  stresses  in  each  section  ? 


THEORY  OF  STRUCTURES. 

(X)  A   trapezoidal  section,  the  top  side,  bottom  side,  and  depth  h 
(inches)  being  in  the  ratio  of  I  to  2  to  4. 

Ans.  %h  from  top  side  ;  TV3<^3  in.-tons. 

(XI)  A  section  in  the  form  of  a  rhombus  of  depth  ic  and  with  a  hori- 
zontal diagonal  of  length  2b.  Ans.  ^bc* ;  ^bc  ;  9  to  8. 

(XII)  An  angle-iron  2  in.  x  2  in.  x  \  in. 

Ans.  Neutral  axis  divides  depth  into  segments  of  J£  in.  and  f| 
in. ;  \\\l  in.-tons ;  ^or  ton  I  !334  to  1369. 

(XIII)  A  hollow  circular  section  of  external  radius  C  and  internal 
radius  C. 

Ans    .99    C*  ~  C'4     33         C4  -  C"  4  C2  +  CC'  +  C " 


112         C        '4   C'2  +  CC'  +  C'  '  3         C'2  +  C'8 
(XIV)  A  cruciform  section  made  up  of  aflat  steel  bar  10  in.  by  i  in. 
and  four  steel  angles,  each  4  in.  by  4  in.  by  £  in.,  all  riveted  together. 
(Neglect  weakening  effect  of  rivet-holes.) 

75.  A  girder  of  21  ft.  span  has  a  section  composed   of  two  equal 
flanges  each  consisting  of  two  3i-in.  x  5~in.  x  i-in.  angles  riveted  to  a 
39-in.  x  f-in.  web  ;  the  cover-plates  on  the  flanges  are  each  12  in.  x  £  in., 
and  the  rivets  in  the  covers  alternate  with  those  connecting  the  angles 
and  web  ;  the  pitch  of  the  rivets  is  3^  in.  Find  the  diameter  and  also  find 
the  maximum  flange  stresses,  (a)  disregard  ing  the  weakening  effect  of  the 
rivet-holes  in  the  tension  flange  ;  (b)  taking  this  effect  into  account, 

The  load  upon  the  girder  is  a  uniformly  distributed  load  of 
20,800  Ibs.  (including  weight  of  girder)  and  a  load  of  50,000  Ibs.  concen- 
trated at  each  of  the  points  distant  4^  ft.  from  the  middle  point  of  the 
girder. 

Ans.  Diam.  of  rivets  =  .48  in.  if  tight,  =  .54  in.  if  subject  to  flexure. 

(a)  fi  =/2  =  7762  Ibs.  per  sq.  in. 

(b)  fi  =  8248  Ibs.  per  sq.  in.,/2  =  7847  Ibs.  per  sq.  in. 

76.  A  beam  of  triangular  section  12  in.  deep  and  with  its  base  hori- 
zontal can  bear  a  total  shear  of  100  tons.     If  the  safe  maximum  intensity 
of  shear  is  4  tons  per  sq.  in.,  find  the  width  of  the  base.       Ans.   6£  in. 

77.  Assuming  that  the  web  and  flanges  of  a  rolled  beam  are  rectangular 
in  section,  determine  the  ratio  of  the  maximum  to  the  average  intensity 

of  shear  in  a  section  from  the  following  data :    the  total  depth  is  —  times 

the  breadth  of  each  flange,  n  times  the  thickness  of  each  flange,  and  2« 
times  the  thickness  of  the  web.  Show  also  that  this  ratio  is  VTB  or  y 
according  as  the  area  of  the  web  is  equal  to  the  joint  area  of  the  two 
flanges  or  is  equal  to  the  area  of  each  flange.  How  much  of  the  shear-- 
ing force  is  borne  by  the  web  ?  How  much  by  the  flange  ? 

3(#2  +  \in  —  \2)(n  +  6) 
Ans.  ratio  =         i-ri-—:  70*1  85*. 


EXAMPLES.  419 

78.  In  a  rolled  beam  with  equal  flanges,  the  area  of  the  web  is  propor- 
tional to  the  nth  power  of  the  depth.  Find  the  most  economical  distribu- 
tion of  metal  between  the  flanges  and  web,  and  the  moment  of  resistance 
to  bending  of  the  section  thus  designed.  Also  find  the  ratio  of  the  aver- 
age to  the  maximum  intensity  of  shear. 

Ans.  Area  of  each  flange  :  web  area  ::  2«  —  i  :  6  ; 


B.  M.  =  - 


2  n 

f  being  the  coefficient  of  strength,  S  the  total  area  of 

section,  and  y  the  depth. 
Max.  intensity  of  shear  :  av.  intensity  ::  (n  +  i)(4«  +  i)  :  6n. 

79.  Find  the  moment  of  resistance  to  bending,  the  resistance  to  shear, 
and  the  ratio  of  maximum  to  the  average  intensity  of  a  shear  in   the 
case  of  a  section   consisting  of  two  equal   flanges,  each  composed  of  a 
pairof  5-in.  x  3^-in.  x  f-in.  angle-irons  riveted  to  a  3ii-in  x  f-in.  web, 
the   5-in.  sides  of  the  angles   being  horizontal,  and  4^  tons  per  sq.  in. 
being  the  coefficient  of  strength. 

Ans.   1501.06  in.  -tons  ;  22.  36  tons;  4.916. 

80.  The  floor-beam  for  a  single-track  bridge  is  15  ft.  between  bearings, 
and  each  of  its  flanges  is  composed  of  a  pair  of  2|-in.  x  2f-in.  x  f-in. 
angle-irons   riveted  to  a  3o-in.  x  f-in.  web.     The  uniformly  distributed 
load  (including  weight  of  beam)  upon  the  beam  is  4200  Ibs.,  and  a  weight 
of  1600  Ibs.  is  concentrated  at  each  of  the  rail-crossings,  2^  ft.  from 
the  centre.     Find  (a)  the  maximum  flange   stress,  (b)  the   ratio  of  the 
maximum  and   average   intensities  of  shear  ;  (c)  the   stiffness,  E  being 
27,000,000  Ibs. 

Ans.  (a)  6523.4  Ibs  ;  (<£)  2.037;  (c)  .00033. 

/  =  •  -  ~  —  ,  neglecting  effect  of  rivet-holes. 

8  1.  A  beam  36  ft.  between  bearings  is  a  hollow  tube  of  rectangular  sec- 
tion and  consists  of  a  24-in.  x  £-in.  top  plate,  a  24-in.  x  £-in.  bottom 
plate,  and  two  side  plates  each  35  in.  x  \  in.  The  plates  are  riveted 
together  at  the  angles  of  the  interior  rectangle  by  means  of  four  6-in. 
x  4-in.  x^-in.  angle-irons,  the  6-in.  side  being  horizontal.  Determine  — 

(a)  The  intensity  of  shear  at  the  surface  between  the  angle-irons  and 
the  upper  and  lower  plates. 

(b)  The  diameter  of  the  rivets,  the  pitch  being  4  in.  and  assuming  an 
effective  width  of  5^  in.  in  shear  per  rivet. 

(c)  The  total  shearing   strength  of  the  section,  the  safe  intensity  of 
shear  being  3^  tons  per  sq.  in. 

(d)  The  moment  of  resistance  of  the  section,  the  coefficient  of  strength 
being  4^  tons  per  sq.  in. 

(e)  The  uniformly  distributed  load  which  the  beam  will  safely  carry. 


42O  THEORY  OF  STRUCTURES. 

Ans. — (a)  .11878  tons  per  sq.  in. 

(b)  .97  in.  if  rivets  are  tight,  1.12  in.  if  liable  to  flexure. 
(<:)  i09Ty^   tons  disregarding  effect  of  riveting, 
tons  having  regard  to  riveting. 

(d)  408  5£in.-tons  disregard  ing  effect  of  riveting,  3838.9157 

tons  having  regard  to  riveting. 

(e)  75f|  tons  disregarding  effect  of  riveting,  71.09   tons 

having  regard  to  riveting. 

82.  A  cast-iron  channel-beam  having  a  web  12  in.  wide  and  two  sides 
7  in.  deep,  the  metal  being  everywhere  i  in.  thick,  crosses  a  span  of  14 
ft.     If  the  tensile  intensity  of  stress  is  i  ton  per  sq.  in.,  what  uniformly 
distributed  load  will  the  beam  carry  (a)  with  the  web  at  the  bottom  ;  (b) 
with  the  web  at  the  top?     Find  (c)  the  maximum  compressive  intensity 
of  stress  to  which  the  metal  is  subjected,  and  (d}  compare  the  maximum 
and  average   intensities  of  shear.     Also,  (e]  what  should  be  the  area  of 
a  rectangular  section  to  bear  the  same  total  shear  ? 

Ans.  7  =  i  io£  ;  (d)  f f|  tons  ;  (ff)  fff  tons. 

83.  A  beam  of  rectangular  section  and  of  a  length  equal  to  20  times  the 
depth  is  supported  at  the  ends  in  a  horizontal  position,  and  is  subjected 
to  a  thrust  //"whose  line  of  action   coincides  with  the  axis  of  the  beam. 
Show  that  the  maximum  intensity  of  stress  at  the  middle  point  will  be 
doubled1  by  concentrating  at  that  point  a  weight  £Fequal  to  one  thirtieth 
of//. 

84.  The  line  of  action  of  the  thrust  in  a  compression   member  is  at  a 
distance   from  the  axis  equal  to  -th  of  the  least  transverse  dimension. 

Show  that  the  maximum  intensity  of  stress  is  doubled   if  the  section  is 
rectangular  and  r  =  6,  or  if  the  section  is  circular  and  r  =  8. 

85.  A  straight  wrought-iron  bar  is  capable  of  sustaining  as  a  strut  a 
weight  Wi,and  as  a  beam  a  weight  w*  at  the  middle  point,  the  deflection 
being  small  as  compared  with  the  transverse  dimensions.     If  the  bar  has 
simultaneously  to  sustain  a  weight  w  as  a  strut  and  a  weight  w'  as  a 
beam,  the  weight  being  placed  at  the  middle  of  the  span,  show  that  the 
beam  will  not  break  if 

w\ 

IV  + W    <  Wi. 

IVl 

86.  A   metal  beam  is  subjected  to  the  action  of  a  bending  moment 
steadily  applied  beyond the  elastic  limit.     Assuming  that  the  metal  acts 
as  if  it  were  perfectly  plastic,  i.e.,  so  that  the  stress  throughout  a  trans- 
verse section  is  uniform,  compare  the  moment  of  resistance  to  bending 
of  a  section  of  the  beam  with  the  moment  on  the  assumption  that  the 
metal  continued  to  fulfil  the  ordinary  laws  of  elasticity,  (a)   the  section, 
being  a  rectangle  ;  (b)  the  section  being  a  circle. 


EXAMPLES.  421 

87.  A  lattice-girder  of  100  ft.  span  carries  80  tons  uniformly  distributed; 
the  girder  is  10  ft.  deep  and  the  safe  working  stress  is  4  tons  per  sq.  in. 
If  the  width  of  the  flange  must  be  20  in.  to  carry  the  load  exclusive  of 
the  weight  of  the  girder,  what  must  be  the  width  of  the  flange  when  the 
weight  of  the  girder  is  taken  into  account  ? 

88.  A  plate-girder  of  double-tee  section  and  of  3o  ft.  span  is  8  ft.  deep 
and  carries  a  uniformly  distributed  load  of  80  tons.     If  the  width  of  the 
flange  must  be  12  in.  to  carry  the  load  exclusive  of  the  weight  of  the  gir- 
der, what  must  the  width  be  when  this  weight  is  taken  into  account  ? 

89.  If  the  plane  of  bending  does  not  coincide  with  the  plane  of  sym- 
metry of  a  beam,  show  that  the  neutral  axis  is  parallel  to  a  line  joining  the 
centres  of  two  circles  into  which  the  beam  would  be  bent  by  two  com- 
ponent couples  whose  axes  are  the  principal  axes  of  inertia  of  the  section, 
each  couple  being  supposed  to  act  alone. 

90.  The  flanges  of  a  girder  are  of  equal  sectional  area,  and  their  joint 
area  is  equal  to  that  of  the  web.    What  must  be  the  sectional  area  to  resist 
a  bending  moment  of  300  in. -tons,  the  effective  depth  being  10  in.  and 
the  limiting  inch-stress  4  tons?  Ans.  22^  sq.  in. 

91.  The  effective  length  and  depth  of  a  cast-iron  girder  which  failed 
under  a  load  of  18  tons  at  the  centre  were  57  in.  and  5^  in.,  respectively ; 
the  top  flange  was  2.33  in.  by  .31  in.,  the  bottom  flange  6.67  in.  by  .66  in., 
and  the  web  was  .266  in.  thick.     Assuming  that  the  ordinary  theory  of 
flexure  held  good,  what  were  the  maximum  intensities  of  stress  in  the 
flanges  at  the  point  of  rupture  ? 

Ans.  ft  =  12.36  tons  per  sq.  in. ;  fc  =  44.9  tons  per  sq.  in. 

92.  A  railway  bridge  is  supported  upon  two  main  girders  each  of  span 
51  ft.  4  in.  ;  at  the  centre  the  depth  is  6  ft.  6  in.,  the  gross  sectional  area 
of  the  top  flange  27  sq.  in.,  and  of  the  bottom  flange  28  sq.  in.     Assum- 
ing the  efficiency  of  the  tension  flange  is  reduced  one-fifth  by  the  rivet- 
holes,  find  the  maximum  flange  intensities  of  stress  under  a  uniformly 
distributed  load  of  43  tons.     Also  find  the  uniformly  distributed  rolling 
load  which  will  increase  these  intensities  by  two  tons. 

Ans.  .786  ton  per  sq.  in.  in  compression  ;  .9475  ton  per  sq.  in.  in 
tension  ;  55/T  tons  to  increase  compression  ;  S9j2^0^  tons  to 
increase  tension. 

93.  A  lattice-girder  of  80  ft.  span  and  8  ft.  deep  is  designed  to  carry  a 
dead  load  of  5o  tons  and  a  live  load  of   120  tons  uniformly  distributed  ; 
at  the  centre  the  net  sectional  area  of  the  bottom  flange  is  45  sq.  in., 
and  the  gross  sectional  area  of  the  top  flange  56^  sq.  in.     Find  the  po- 
sition of  the  neutral  axis  and  the  maximum  flange  intensities  of  stress. 
If  the  live  load  travels  at  60  miles  an  hour,  what  will  be  the  increased 
pressure  due  to  centrifugal  force? 


422  THEORY  OF  STRUCTURES. 

Ans.  3.546  ft.  from  top;    1120  Ibs.  per  sq.  in.;  8920.35  Ibs.  per 

594000 
sq.  ID.  ;  — -pj —  Ibs. 

94.  Determine  the  thickness  of  the  metal  in  a  cast-iron  beam  of  12  ft. 
span  and  8  in.  deep  which  has  to  carry  a  uniformly  distributed  load  of 
4000  Ibs.,  the   section    being  (a)  a  hollow  square  ;  (b)  a  circular  annulus. 
The  coefficient  of  working  strength  =  3000  Ibs.  per  sq.  in.    Also  find  the 
limiting  safe  span  of  the  beam  under  its  own  weight. 

Ans.  Neglecting  weight  of  beam,  (a)  .281  in.;  (b)  477  in.  Taking 
weight  of  beam  into  account,  (a)  .307  in.  ;  (b)  .534  in.  Limiting 
span  =  41.3  ft.  in  (a)  and  =  35.7  ft.  in  (b). 

95.  Determine    suitable  dimensions  for  a  cast-iron  beam  20  in.  deep, 
at  a  section  subjected  to  a  bending  moment  of  1200  in.-tons  ;  the  coeffi- 
cients of  strength  per  square  inch  being  2  tons  for  tension  and  8  tons  for 
compression.     Take  thickness  of  web  =  T9^  in. 

Ans.  Sectional  area  of  tension  flange  =  36  sq.  in.;  of  compression 
flange  =  2j  sq.  in. 

96.  The  thickness  of  the  web  of  an  equal-flanged  I-beam  is  a  certain 
fraction  of  the  depth.     Show  that  the  greatest  economy  of  material  is 
realized  when   the   area  of  the  web  is   equal  to  the  joint  area  of  the 
flanges,  and  that  the  moment  of  resistance  to  bending  is  ^/Ad,f  being 
the  coefficient  of  strength,  A  the  total  sectional  area,  and  d  the  depth. 

97.  In  a  double-flanged  cast-iron  beam  the  thickness  of  the  web  is  a 
certain  fraction  of  the  depth,  and  the  maximum  tensile  and  compressive 
intensities  of  stress  are  in  the  ratio  of  2  to  5.     Show  that  the  greatest 
economy  of  material  is  realized  when  the  areas  of  the  bottom  flange,  web, 
and  top  flange  are  in  the  ratio  of  25  to  20  to  4,  and  that  the  moment  of 
resistance  to  bending  is  \fAd,  where/ =  -^  maximum  tensile  intensity 
of  stress. 

98.  Apply  the  results  in  the  preceding  question  to  determine  the  di- 
mensions of  a  cast-iron  beam  at  a  section  whose  moment  of  resistance  is 
800  in.-tons  and  whose  depth  is  18  in.,  taking  2  tons  per   square  inch  as 
the  maximum  tensile  intensity  of  stress. 

Ans.  ai  =  ^fi-  sq.  in.;  A'  =  -4^  sq.  in.;  a^=  |^  sq.  in. 

99.  Determine  suitable  dimensions  for  a  cast-iron  girder  of  20  ft.  span 
and   24    in.    deep,    carrying  a  load   of  30,000  Ibs.    at   the   centre,   the 
coefficients  of  working  strength  in  tension  and  compression  being  respec- 
tively 2000  and  5000  Ibs.  per  square  inch. 

Ans.  ai  =  ijf1  sq.  in. ;  A'  =  &3^  sq.  in.  ;  a*  =  ty  sq.  in. 

100.  A  cast-iron  girder  of  25  ft.  span  has  a  bottom  flange  of  36  sq.  in. 
sectional  area.  Find  the  most  economic  arrangement  of  material  for  the 
web  and  top  flange  which  will  enable  the  beam  to  carry  a  load  of  18,900 
Ibs.  at  IG  ft.  from  one  end. 


EXAMPLES.  423 

Am.  Depth  =  2oJ  in.;  area  of  web  =  28.8  sq.  in.;    area  of  top 
flange  —  5.76  sq.  in. 

101.  A  double-flanged  cast-iron  girder  has  a  sectional  area  of  93  sq. 
in. ;  the  web  is  i  in.  thick  and  21  in.  deep ;  the  moment  of  resistance  of 
the  section  is  100,950  ft.-lbs. ;  the  coefficients  of  strength  are  2100  Ibs. 
per  square  inch   in  tension  and   5250  Ibs.    in  compression.     Find  the 
position  of  the  neutral  axis  and  the  areas  of  the  two  flanges. 

102.  Determine  the  moment  of  resistance  to  bending  of  a  section 
of  a  beam  in  which  the  top  flange  is  composed  of  two  34o-mm.  x  12-mm. 
plates  and  one  34o-mm.  x  lo-mm.  plate,  and-  the  bottom  flange  of  one 
340- mm.  x  lo-mm.   plate  and  one   34o-mm.  x  8-mm.  plate,  the  flanges 
being    riveted    to    a     i.4-m.  x  7-mm.    web    plate    by   means   of    four 
loo-mm.  x  loo-rnm.  x  8-mm.   angle-irons.     The  coefficient  of  strength 
=  6  k.  per  mm.2. 

103.  Compare  the  moments  of  resistance  to  bending  of  the  section  in 
the  preceding  question  and  of  a  section  in  which  three  4oo-mm.  x  15-mm. 
plates  are  substituted  for  the  top  flange,  and  one  4oo-mm.  x  i5-mm. 
plate  is  substituted  for  the  bottom  flange. 

104.  Floor-beams  4.4m.  between  bearings  and  spaced  2.548  m.  c.  to  c. 
have  a  section  composed  of  two  equal  flanges,  each  consisting  of  two 
85-mm.  x  85-mm.  x  i2-mm.  angle-irons  riveted  to  a  49o-mm.  x  7-mm. 
web.    A  weight  of  1 50  k.  (due  to  longitudinals)  and  a  weight  of  1 50  k.  (due 
to  rails, etc.),  i.e.,  300  k.  in  all,  are  concentrated  at  the  rail-crossings,  and 
the  ties  have  also  to  carry  a  uniformly  distributed  load  of  400  k.  due  to 
weight  of  floor-beam,  4000  k.  due  to  weight  of  platform,  and  4000  k.  per 
square  metre  of  platform  due  to  proof-load.     Find  the  moment  of  resist- 
ance to  bending  and  the  maximum  flange  intensities  of  stress. 

Ans.  /=  .000438584615. 

105.  The  section  of  a  beam  is  in  the  form  of  an   isosceles  triangle 
with    its   base   horizontal.     Show   that   the  moment    of    resistance   to 
bending  of  the  strongest  trapezoidal  beam  that  can  be  cut  from  it  is 
very  nearly  -^^fbd^,  b  being  the  width  of  the  base  and  d  the  depth  of 
the  triangle. 

106.  Taking/f,/c  as  the  tensile  and  compressive  intensities  of  stress, 
find  the  moment  of  resistance  to  bending  of  a  section  consisting  of  a 
2o/-in.  x  7^-in.   top   flange,  an   8o/-in.  x  io/-in.    bottom    flange,    and   a 
trapezoidal  web  4/  in.  thick  at  the  top,  8/  in.  thick  at  the  bottom,  and 
I20/  in.  deep.     Also  compare  the  maximum  and  average  intensities  of 
shear. 

107.  Each  of  the  flanges  of  a  girder  is  a  35o-mm.  by  lo-mm.  plate  and 
is   riveted   to   a    i.8-m.  by  8-mm.  web   by   means   of  two   loo-mm.  by 
loo-mm.  by  12-mm.  angle-irons.     Determine  the  moment  of  resistance 
to  bending,  the  coefficient  of  strength  being  6k.  per  square  millimetre, 


424  THEORY  OF  STRUCTURES. 

(a)  disregarding  the  weakening  effect  of  riveting ;  (b)  assuming  that  the 
flange-plates  are  riveted  to  the  angles  by  2o-mm.  rivets. 

Ans.   (a)  108661.04  km.     ' 

108.  The  cross-tie  for  a  single-track  bridge  is  4.1  m.  between  bearings, 
the  gauge  of  the  rails  being  1.51  m. ;  each  of  the  flanges  is  composed  of  a 
148-mm.  by  8-mm.  plate  riveted  to  a  55o-mm.  by  8-mm.  web  by  means  of 
two  7o-mm.  by  /o-mm.  by  9-mm.  angle-irons;  a  load  of  296  k.   (weight 
of  rails,  etc.)   is   concentrated  at  each  rail-crossing.     What    uniformly 
distributed  load  will  the  tie  safely  bear,  the  metal's  coefficient  of  strength 
being  6  k.  per  square  millimetre  ?    The  load  actually  distributed  over  the 
tie  is  19782  k.     Find  the  maximum  intensity  of  stress. 

Ans.  24162  k.  ;  4.94  k.  per  sq.  mm. 

109.  Design  a  longitudinal  of  .45  m.  depth  which  is  to  be  supported 
at  intervals  of  3.3  m.  and  to  carry  at  its  middle  point  a  weight  of  7000  k., 
the  coefficient  of  strength  being  5  k.  per  square  millimeter. 

Ans.  I  =  259.875,  and  the  /of  a  section  with  two  equal  flanges, 

each  composed  of  two  7o-mm.  by  7o-mm.  by  9-mm.  angle-irons 

riveted  to  a  45o-mm.  by  8-mm.  web  is  259.102455. 

no.  Find  the  moment  of  resistance  of  a  section  composed  of  two 

equal   flanges,  each   consisting  of  two  6oo-mm.  x  7-mm.  plates  riveted 

to  a  i2OO-mm.  x  8-mm.  web  plate  by  means  of  two  loo-mm.  x  loo-mm. 

x  i2-mm.  angle-irons;  two  7o-mm.  x  7o-mm.  x  9-mm.  angles  are  also 

riveted  to  the  lower  faces  of  the  flanges,  the  ends  of  the  horizontal  arms 

being  24  mm.  from  the  outside  edges  of  the  flanges ;  the  total  depth  of 

the  section  =  3.228  m.,  and  the  interval  between   the   two  web  plates, 

which  is  open,  is  2  m. ;  coefficient  of  strength  =  6  k.  per  mm.'2. 

Ans.  /=  .093929232444  and  moment  =  349179.3018  km. 
in.  A  longitudinal  2.548  m.  between  bearings  consists  of  two  equal 
flanges,  each  composed  of  two  7o-mm.  x  7o-mm.  x  9-mm.  angle-irons 
riveted  to  a  35o-mm.  x  7-mm.  web  plate.     Find  the  flange  intensity  of 
stress  under  a  maximum  load  of  7000  k.  at  the  centre. 

Ans.  I  =  .000139284508;  stress  =  5.6  k.  per  mm.2. 
1 1 2.  A  cross-tie  resting  upon  supports  at  the  ends  and  2.26  m.  between 
bearings  is  composed  of  two  equal  flanges,  consisting  of  two  7o-mm. 
x  7o-mm.X9-mm.  angle-irons  riveted  at  the  top  to  a  45o-mm.  x7-mm. 
web  plate  and  at  the  bottom  to  a  3oo-mm.  x  7-mm.  web  plate,  the 
interval  between  the  web  plates,  which  is  open,  being  2.55  m. ;  the  tie  is 
designed  to  carry  a  uniformly  distributed  load  of  676  k.  per  lineal  metre 
of  its  length,  and  also  a  load  of  11644.8  k.  at  each  of  the  points  distant 
.375  m.  from  the  bearing.  Find  the  position  of  the  neutral  axis  and  the 
maximum  flange  stresses. 

Ans.  1.516  m.  from  top  flange;  7=  .023194564198;  maximum 
B.  M.  =  4815.8161  km.  ;  maximum  tensile  stress  =  .37  k.  per 
mm.2 ;  maximum  compressive  stress  =  .314  k.  per  mm.2. 


EXAMPLES.  425 

113.  Find  the  maximum  concentrated  load  on  a  cross-tie  for  a  single 
track  due  to  a  six-wheel  locomotive,  the  wheels  being  2.3  m.  centre  to 
centre,  the  ties  being  3.2  m.  centre  to  centre,  and  the  weight  on  each 
wheel  being  7000  k.  Ans.  10937.5  k. 

114.  The  floor-beams  for  a  double-track   bridge  are  8.3m.   between 
bearings  and  are  spaced  2.58  m.  centre  to  centre.     The  distance,  centre 
to  centre,  between  track-rails  is  1.5  m.,  and  between  inside  rails  is  2  m.  ; 
the  tie  has  equal  flanges,  each  consisting  of  two  7o-mm.  x  7o-mm.  x  9-mm. 
angle-irons    riveted   to  a  6oo-mm.  x  7-mm.  web;    the  maximum  live 
load  upon  the  tie  is  that  due  to  a  weight  of  7000  k.  upon  each  of  the  six 
wheels  of  two  locomotives,  the  wheels  being  2.4  m.  centre  to  centre.     If 
the  coefficient  of  working  strength  is  5!  k.  per  square  millimetre,  what 
uniformly  distributed  load  will  the  tie  carry? 

115.  Determine   the  safe  value  of  the   moment  of  inertia  (/)  of  a 
cross-tie  for  a  double-track  bridge  ;  the  length  of  the  tie  between  bearings 
being  7. 624  m.,  its  depth  .6  m.,the  gauge  of  the  rails  1.5  m.,  the  distance 
between    inside  rails  2  m.     The  uniformly  distributed   load  upon  a  tie 
consists  of  850  k.  per  square  metre  due  to  platform,  etc.,  and  of  1800  k. 
due  to  weight  of  tie  ;  the  ties  are  3.584  m.  centre  to  centre  ;  the  live  load 
is  that  due  to  a  weight  of  7000  k.  upon  each  of  the  centre  wheels  of  a  six- 
wheel  locomotive  and  a  weight  of  6000  k.  upon  each  of  the  front  and  rear 
wheels,  the  wheels  being  2.4  m.  centre  to  centre;  the  safe  coefficient  of 
strength  =  6  k.  per  square  millimetre. 

116.  The  upper  chord  of  a  Howe  truss  is  24  in.  wide  x  12  in.  deep 
and  is  made  up  of  four  12-in.  x  6-in.  timbers;  the  lower  chord  is  24  in. 
wide  x  16  in.  deep  and  is  made  up  of  four  i6-in.  x  6-in.  timbers  ;    the 
distance  between  the  inner   faces  of  the  chords  is  24  ft.      Find  the  mo- 
ment of  resistance  to  bending,  taking  800  Ibs.  per  square  inch  as  the  co- 
efficient of  tensile  strength,  and  neglecting  the  effect  of  the  web. 

Ans.  Neutral  axis  is  137^  in.  from  bottom  face  of  lower  chord  ; 
moment  =  87441616  in. -Ibs. 

117.  The  cross-ties  of  a  single-track  bridge   consist  of  two   equal 
flanges,  each  composed  of  two  7o-mm.   x  70  mm.  x  9  mm.  angle-irons 
riveted  to  a  65o-mm.  x  7-mm.  web;   the  ties  are  4.1  m.  long,  and  each 
carries  19,146  k.  (viz.,  384  k.for  ties,  2762  k.for  platform,  and  16,000  k.for 
proof  load}  uniformly  distributed  and  635  k.  (due  to  longitudinals,  rails, 
etc}  concentrated  at  each  rail-crossing,  i.e.,  at  755  mm.  from  the  middle 
point.     Assuming  that  the  cross-ties  are  merely  supported  at  the  ends, 
find  the  maximum  intensity  of  stress. 

Ans.  5.7724  k.  per  mm.2 ;  — =  .0018423. 
N.B. — The  fixture  of  the  ends  approximately  doubles  the  strength. 


426  THEORY  OF  STRUCTURES. 

1 1 8.  The  longitudinals  of  the   bridge  in  the  last  question  consist  of 
two  pairs  of   7o-mm.  x  7o-mm.  x  g-mm.  angle-irons   riveted  to  a  4m. 
x  7  mm.  web ;  the  cross-ties  are  3.2  m.  centre  to  centre.     Determine  the 
maximum  intensity  of  stress  due  to  a  load  of  7000  k.  concentrated  on  the 
longitudinal  half-way  between  the  cross-ties,  assuming  that  it  is  an  inde- 
pendent girder.    What  would  the  stress  be  if  the  ties  were  3  m.  centre  to 

centre  ?  / 

Ans.  —  =  .00095458;  5.866k.  per  mm.2;  5. 4994k.  per  mm.2 

119.  The  section  for  the  Estressol  bridge  cross-ties  is  the  same  as  that 
for  the  Grande  Baise  (Ex.  117)  bridge  ties  ;  the  load  at  each  rail-crossing 
is  335  k.,  and  the  uniformly  distributed  load  is  18,062  k.     Find  the  max- 
imum  intensity  of  stress  in   the  flanges,   assuming  that   the  ties  are 
merely  supported  at  the  ends.  Ans.  5.26  k.  per  mm.2 

120.  In  a  rolled  joist  the  sum  of  the  two  flange   areas  and  the  web 
area  is  a  constant  quantity.     Find  the  proportion  between  them  which 
will  give  a  joist  of  maximum  strength,  the  thickness   of  the   web  being 
fixed  by  practical  considerations.          Ans.  Flange  area  =  f  web  area. 

121.  An  aqueduct  for  a  span  of  26  ft.  consists  of  a  cast-iron  channel- 
beam  30  in.  wide  and  20  in.  deep.     Find  the  thickness  of  the  metal  so 
that  the  water  may  safely  rise  to  the  top  of  the  channel,  the  safe  coeffi- 
cient of  strength  being  i  ton  per  square  inch.     Find  the  safe  limiting 
span  of  the  channel  under  its  own  weight. 

122.  A  rolled   beam  with  equal  flanges  and  a  web  whose  section  is 
equal  to  the  joint  section  of  the  flanges  has  a  span  of  24  ft.  and  carries  a 
weight  of  8  tons  at  the  centre.    If  the  stiffness  is  .001  and  if  the  coefficient 
of  strength  per  square  inch  is  5  tons,  find  the  depth  of  the  beam  and  the 
web  and  flange  sectional  areas. 

123.  A   wrought-iron    beam   of   I-section,  20   ft.  between   supports, 
carries  a  uniformly  distributed  load  of  4000  Ibs.  and  deflects  .1  in.;  the 
effective  depth  =8  in.;  .£"=30,000,000  Ibs.;    web  area  =  joint  area  of 
the  equal  flanges.    Find  the  total  sectional  area.    Also  find  the  width  of 
a  rectangular  section  8  in.  deep  which   might  be  substituted  for   the 
above.  Ans.  /=  288  ;  area  =  27  sq.  in.;  width  =  6f  in. 

124.  A  cast-iron  beam  of  an  inverted  T-section  has  a  uniform  depth 
of  20  in.  and  is  22  ft.  between  supports;  the  flange  is  12  in.  wide  and  1.2 
in.  thick  ;  the  web  is  i  in.  thick  ;  the  load  upon  the  beam  is  4500  Ibs. 
per  lineal  foot;  £=  17,000,000  Ibs.     Find  the  deflection  at  the  centre, 
the  moment  of  resistance  to  bending,  the   maximum   tensile  and  com- 
pressive  intensities  of  stress,  and  the  position  of  the  neutral  axis.     Why 
is  the  flange  placed  downwards  ? 

125.  Find  the  sectional  area  of  a  wrought-iron   beam  of  T-section 
which  may  be  substituted  for  the  cast-iron  beam  in  the  preceding  ques- 
tion, the  depth  being  the  same  and    the   coefficients  of   strength    per 


EXAMPLES.  427 

square  inch  being  3  tons  in  compression  and  5  tons  in  tension.  Why 
should  the  flange  be  uppermost  ?  What  should  the  total  sectional  area 
be  if  the  flange  and  web  are  of  equal  area  ? 

126.  A  cast-iron  girder  139  in.  between  supports  and  10  in.  deep  had 
a  top  flange  2^  in.  x  £•  in.,  a  bottom  flange  10  in.  x  i  J  in.,  and  a  web  f  in. 
thick.     The  girder  failed   under  loads  of   17^  tons  placed  at  the  two 
points  distant  3!  ft.  from  each  support.     What  were  the  central  flange 
stresses  at  the  moment  of  rupture  ?     What  was  the   central  deflection 
when  the  load  at  each  point  was  7!  tons  ?     (E  =  18,000,000  Ibs.  ;  weight 
of  girder  =  3368  Ibs.;  ton  =  2240  Ibs.) 

Ans.  182251.9  Ibs.  =  total   flange    stress;    unit  flange  stresses 
=  14,580,  and  41,657  Ibs.  persq.  in.;  deflection  =  .35". 

127.  A  cylindrical  beam  of  2  in.  diameter,  60  in.  in  length,  and  weigh- 
ing \  Ib.  per  cubic  inch,  deflects  T%  in.  under  a  weight  of  3000  Ibs.  at  the 
centre.     Find  E.  Ans.  E=  21,645,511  Ibs. 


CHAPTER   VII. 
ON   THE   TRANSVERSE   STRENGTH    OF    BEAMS.— Continue*. 

I.  General  Equations. — The  girder  OA  of  length  /  carries 
a  load  of  which  the  intensity  varies  continuously  and  is  /  at  a 
point  K  distant  x  from  O. 

M+dM 
K    V  "* 


V 

FIG.  310. 

Consider  the  conditions  of  equilibrium  of  a  slice  of  the 
girder  bounded  by  the  vertical  planes  KL,  K  'L  ',  of  which  the 
abscissae  are  x,  x  +  dx,  respectively. 

The  load  between  these  planes  may,  without  sensible  error, 
be  supposed  to  be  uniformly  distributed,  and  its  resultant  pdx 
therefore  acts  along  the  centre  line  VV  . 

The  forces  acting  upon  the  slice  at  the  plane  KL  are  equiv- 
alent to  an  upward  shearing  force  S,  and  a  right-handed  couple 
of  which  the  moment  is  M,  while  the  forces  acting  upon  the 
slice  at  the  plane  K'  L'  are  equivalent  to  a  downward  shearing 
force  5  +  dSt  and  a  left-handed  couple  of  which  the  moment 


Since  there  is  to  be  equilibrium, 
S—(S~\-dS)—pdx=  the  algebraic  sum  of  the  vertical  forces=o. 


428 


GENERAL   EQUATIONS.  429 

And,  M-(M+JM)  +  S—  +  (S+JS)^=  the   alge- 
braic sum  of  the  moments  of  the  forces  with  respect  to  V  or 


The  term  —      -  is  disregarded,  being  indefinitely  small  as 

compared  with  the  remaining  terms. 

Equations  (a)  and  (b)  are  the  general  equations  applicable 
to  girders  carrying  loads  of  which  the  intensity  is  constant  or 
varies  continuously.  Their  integration  is  easy,  and  introduces 
two  arbitrary  constants  which  are  to  be  determined  in  each 
particular  case. 

Cor.  i.  From  equations  (a)  and  (3), 

_dS_ 
*  ~~~        ~       P' 


dx*       dx 

Let  p  =  wf(x),  w  being  a  constant,  and  f(x)  some  function 
of  x.     Then 

dM 


and 

M=c, 


cl  and  £,  being  the  constants  of  integration,  and  o  and  x  the 
limits. 

EXAMPLE.  —  Let  the  girder  rest   upon  two  supports   and 
carry  a  uniformly  distributed  load  of  intensity  w^     Then 


and 

M  =  c^  -[-  cjc  —  wl  —  . 


43°  THEORY   OF  STRUCTURES. 

But  M  is  zero  when  x  =  o  and  also  when  x  =  /.     Hence 


Therefore, 


and 


a  =  o     and     ct  =  — . 


,  l 

M  —  —x -x\ 

2  2 


-j—  — W.X. 

dx          2 


.  2.  The  bending  moment  is  a  maximum  at  the  point 

/V  7l/f 

defined  by  -7—  =  o  =  5,  i.e.,  at  a  point  at  which  the  shearing 

force  vanishes. 

In  the  preceding  example,  the  position  of  the   maximum 

bending  moment  is  given  by  S  —  o  '=  — w^,    or  x  =  — , 

.  wj  /    w.  r    w.r 

and  its  corresponding  value  is =  ~^~- 

22  24  o 

W  / 

The  shearing  force  is  greatest  and  equal  to  —  when;r=O. 

Cor.  3.  Suppose  that  the  load,   instead  of  varying  contin- 
Nr  Nr-fi     uous^y»  consists  of  a  number  of  finite  weights 

at  isolated  points. 

By  reason  of  the  discontinuity  of  the  load- 


FIG.  3n.  mg^  j-^e  general  equations  can  only  be  inte- 

grated between  consecutive  points. 

Let  Nrj  Nr+l  ,  be  any  two  such  points,  of  abscissae  xr  ',  xr+l  , 
respectively. 

Between  these  points  equations  (a)  and  (£),become 

dS  dM 


/.  5"  =  a  constant  =  S,  ,  suppose,  between  Nr  and  Nr+l  . 


GENERAL   EQUATIONS.  431 

Hence,   -j-    =  5r,  and   ^f=Sr^+  £,   between   Nr  and 

_,_,  ,  £  being  a  constant  of  integration. 
Let  M  =  .Afr  when  ;tr  =  ^rr.     Then 


c  =  Mr  —  SrXr ,    and     J/  =  Sr(;r  -  #r 
Also,  if  M  —  Mr+l  when  x  =  xr+^ , 


The  terminal  conditions  will  give  additional  equations,  by 
means  of  which  the  solution  may  be  completed. 

EXAMPLE. — The  girder  OA,  of  length  /,  rests  upon  two 
supports  at  0,  A,  and  carries  weights  Plt  /*,,  at  points  B,  C, 


R? 


Vl 


FIG.  312. 

dividing  the  girder  into  three  segments,  OB,  BC,  CA,  of  which 
the  lengths  are  r,  s,  t,  respectively. 


The  reaction  R,  at  O  = 


The  reaction  R,  at  A  =  P*r  +  Pj(r  +  s\ 

Between  O  and  B,  S  is  constant  =  Sr  suppose,  =  Rt  . 


there  being  no  constant  of  integration,  as  M  =  o  when  x  =  o. 
Also,  when  x  =  r,  J/  *=  .Srr. 


43  2  THEORY  OF  STRUCTURES. 

Between  B  and  C,  S  is  constant  =  5f    suppose,  =  Rt  —  Pv 


cf  being  the  constant  of  integration. 
But  M  =  Srr  when  x  =  r. 


and     M  =  S>  +  (Sr  -  S,)r 


Also,  when  x  =  r  -{-  s,  M  =  Sss  -\-  Srr. 
Between  CandA,  S  is  constant  =  St  suppose,  = 
and  hence 


c"  being  the  constant  of  integration. 
But  M  =  SsS  -\-  Srr  when  x  =  r  +  s. 

.-.  c"  =  Sj  +  Srr  -  Sjr  +  s), 
and 

M=  Ss  +  5>  +  Srr  -  St(r  -f  s). 


Hence,  at  At  o  =  Stt  +  5>  +  5>. 

6^r.  4.    The  equation  -—  -  =  5  indicates  that  the  shearing 

force  at  a  vertical  section  of  a  girder  is  the  increment  of  the 
bending  moment  at  that  section  per  unit  of  length,  and  is  an 
important  relation  in  calculating  the  number  of  rivets  required 
for  flange  and  web  connections. 

2.  On  the  Interpretation  of  the  General  Equations.  — 
The  bending  moment  M  at  any  transverse  section  of  a  girder 

El 

may   be  obtained  from  the   equation  M=  —  ,R  being  the 

•  .A. 


DEFLECTION  OF  BEAMS.  433 

radius  of  curvature  of  the  neutral  axis  at  the  section  under 
consideration.  \ 

Let  OA,  in  Figs.  313  and  314,  represent  a  portion  of  the 
neutral  axis  of  a  bent  girder. 


--^  Jy 


FIG.  313.  FIG.  314. 

Take  O  as  the  origin,  the  horizontal  line  OX  as  the  axis  of 
x,  and  the  line  O  Y  drawn  vertically  downwards  as  the  axis  of  y. 

Let  x,  y  be  the  co-ordinates  of  any  point  P  in  the  neutral 
axis. 

If  R  is  the  radius  of  curvature  at  P,  then 


dx*  a  d& 

" 


the  sign  being  -f-  or  —  according  as  the  girder  is  bent  as  in 
Fig.  313  or  as  in  Fig.  314,  and  0  being  the  angle  between  the 
tangent  at  P  and  OX. 

dy 
Now,  jk  is  the  tangent  of  the  angle  which  the  tangent  line 

at  P  to  the  neutral  axis  makes  with  OX,  and  the  angle  is  always 
very  small.  Thus,  -j-  is  also  very  small,  and  squares  and 

dy 

higher  powers  of  -7—  may  be  disregarded  without  serious  error. 

Hence, 


I  ~  , 

•„  =  ±  -jTT»   approximately, 

and  the  bending-moment  equation  becomes 
M  =  ±  EI-jj. 


434  THEORY  OF  STRUCTURES. 

The  integration  of  this  equation  introduces  two  arbitrary 
constants,  of  which  the  values  are  to  be  determined  from  given 
conditions.  At  the  point  or  points  of  support,  for  example, 
the  neutral  axis  may  be  horizontal  or  may  slope  at  a  given 
angle. 

Let  6  be  the  slope  at  P.     Since  6  is  generally  very  small, 

6  —  tan  6,  approximately, 
and  hence 


or 

dB  M 


From  this  last  equation 
A"  0= 


and  the  change  of  slope  between  any  given  limits  is  represented 
by  the  corresponding  area  of  the  bending-moment  curve. 

dy  r 

Also,  since  —  =  0,       y  =  J  6dx, 

d>& 

and  the  deflection  is  measured  by  the  area  of  a  curve  repre- 
senting the  slope  at  each  point. 
Again,  by  Art.  i, 


Comparing   eqs.  (A)   and    (B),    it   will   be   observed    that 

M 
y,  Oj  and  -^-,,  i.e.,  the  deflection,  slope,  and  bending  moment, 

are  connected  with  one  another  in  precisely  the  same  manner 


NEUTRAL   AXIS   OF  A    LOADED  BEAM.  435 

as  M,  S,  and  p,  i.e.,  the  bending  moment,  shearing  force,  and 
load.  Thus,  the  mutual  relations  between  curves  drawn  to 
represent  the  deflection,  slope,  and  bending  moment  must  be  the 
same,  mutatis  mutandis,  as  those  between  the  curves  of  bending 
moment,  shearing  force,  and  load. 

For  example,  divide  the  effective  bending-moment  area  into 
a  number  of  elementary  areas  by  drawing  vertical  lines  at  con- 
venient distances  apart,  and  suppose  these  elementary  areas  to 
represent  weights.  Two  reciprocal  figures  connecting  y,  0,  and 
M  may  now  be  drawn  exactly  as  described  in  Chap.  I,  and  it 
at  once  follows  that  — 

(a)  Any  two  sides  of  the  funicular  polygon,  or,  in  the  limit 
(when   the   widths   of    the   elementary  areas   are   indefinitely 
diminished),  any  two  tangents   to  the  funicular   or  deflection 
curve,  meet  in  a  point  which  is  vertically  below  the  centre  of 
gravity  of  the  corresponding  effective  moment  area. 

(b)  The  segments  iH,  nH  into  which  the  line  of  weights  is 
divided  by  drawing  OH  parallel  to  the  closing  line  CD,  give  the 
slopes  (=  ~2Mdx)  at  the  supports. 

N.B.  —  In  the  case  of  a  semi-girder,  the  last  side  of  a 
polygon  is  the  closing  line,  and  in  gives  the  total  change  of 
slope. 

(c)  If  the  polar  distance  is  made  equal  to  El,  the  intercept 
between  the  closing  line  and  the  funicular  or  deflection  curve 
measures  the  deflection. 

3.  Examples  of  the  Form  assumed  by  the  Neutral 
Axis  of  a  Loaded  Beam. 

EXAMPLE  I.  A  semi-girder  fixed  at  one  end  O  so  that  the 
neutral  axis  at  that  point  is  horizontal  carries  a  weight  W  at 
the  other  end  A.  At  any  point  (x,  y)  of  the  neutral  axis 


x).   .    .    (A) 
Integrating, 


£l  being  a  constant  of  integration.    But  FlG-  3*s- 

the  girder  is  fixed  at  O,  so  that  the  inclination  of  the  neutral 


THEORY  OF  STRUCTURES. 


axis  to  the  horizon  at  this  point  is  zero,  and  thus,  when  x  =  o, 

dy 

-j-  is  o,  and  therefore  ct  =  o. 

Hence, 

%:  ^4-D  .....    .     .     (B 

Integrating, 


£2  being  a  constant  of  integration.      But  y  =  o  when  x  =  o, 
and  therefore  £a  =  o. 
Hence, 


dy 
Equation  (B)  gives  the  value  of  —  ,  i.e.,  the  slope,  at  any 

point  of  which  the  abscissa  is  x. 

Equation  (C)  defines  the  curve  assumed  by  the  neutral  axis, 
and  gives  the  value  of  y,  i.e.,  the  deflection,  corresponding  to 
any  abscissa  x. 

Let  <*!  be  the  slope,  and  dl  the  deflection  at  A. 

From  (B), 


w 


and  from  (C), 


i  w  r 

d ,  — ^~  T- 


NEUTRAL  AXIS  OF  A   LOADED  BEAM. 


437 


Ex.  2.  A  semi-girder  fixed  at  one  end  O  carries  a  uniformly 
distributed  load  of  intensity  w. 

At   any  point   P  (x,  y)    of    the 
neutral  axis, 


.    .    (A)    Y 


FIG.  316. 


Integrating, 


='•*  -<*+$+>.. 


l  being  a  constant  of  integration. 

dy 
But  -j-  =  o  when  x  =  o,  and  therefore  ^  =  O.     Hence, 


dx 


(B) 


Integrating, 


a  being  a  constant  of  integration. 

But  j  =  o  when  x  =  o,  and  therefore  £2  =  O.     Hence, 


2          2 


(Q 


Let  «a  be  the  slope  and  d^  the  deflection  at  A.     Hence,  from 


(B), 


and  from  (C), 


i  wr 


43$  THEORY  OF  STRUCTURES. 

Ex.  3.  A  semi-girder  fixed  at  one  end  carries  a  uniformly 
distributed  load  of  intensity  w,  and  also  a  single  weight  Wat  the 
free  end.  This  is  merely  a  combination  of  Examples  I  and  2, 
and  the  resulting  equations  are  : 

x)  +      (l-xf (A) 


Also,  if  A  is  the  slope  and  D  the  deflection  at  the  free  end, — 
from  (B), 

i  iwr    wi\ 

tan  A  =  -g^ 1-  -^-  ]  =  tan  or,  -j-  tan  <*a ; 

and  from  (C), 


. — The  slope  (a)  and  deflection  (</)  of  an  arbitrarily 
loaded  semi-girder  may  be  determined  in  the  manner  de- 
scribed in  Art.  2. 

Let  F  be  the  area  of  the  bending-moment  curve.  Its 
centre  of  gravity  is  at  the  same  horizontal  distance  ^  from  the 
vertical  through  A  as  the  point  T  in  which  the  tangent  at  A 
intersects  OX. 

.'.  -£-T=  <*  =  angle  A  TX  =  — . 

Jj*  J.  3v 

In  Ex.  3,  e.g., 

F=mt  +  L^ 

2  32 

and 


NEUTRAL  AXIS   OF  A    LOADED  BEAM.  439 

Note.  —  If  the  semi-girder  in  the  three  preceding  examples 
is  only  partially  fixed  at  O,  so  that  the  neutral  axis,  instead  of 
being  horizontal  at  the  support,  slopes  at  an  angle  0,  then 

dy 
when  x  =  o,  -y-  =  tan  6,  and  the  constant  of  integration, 

£,  ,  is  also  El  tan  0.     Thus,  the  left-hand  side  of  eqs.  (B)  and 
(C),  respectively,  become 


\alc  ~ 


EI         ~  tan  B      and 


Ex.    4.     The    girder  OA    rests  upon  two  supports  at    <9, 
A,  and  carries  a  weight  W  at  the  B 

centre.  °r^-  \V      i           ~       * 

The  neutral    axis   is   evidently  p      J    " 

symmetrical   with    respect    to   the  |y                w 

middle  point   C,  and  at  any  point  FlG>  3I7' 
P  (x,  y)  between  O  and  C, 

EI<fy=w_x  (A} 

dx*  ~'   2  ^    ' 

Integrating, 

~EI~dlc"~'~^*i  +  c^ 

^  being  a  constant  of  integration. 

But  the  tangent  to  the  neutral  axis  at  C  must  be  horizontal, 

/   dy  Wr 

so  that  when  x  —  -,  -y-  —  o,  and  therefore  c.  = ^-  . 

2   dx  16 

Hence, 

~EI~d~x=~^X*       l6~ (B) 

Integrating, 

—  Ely     =  — x* —x  -4-  cn . 

12  16 

c9  being  a  constant  of  integration. 


44°  THEORY  OF  STRUCTURES. 

But  y  =  O  when  x  =  o,  and  therefore  £a  =  o.     Hence 


(Q 


Cor. — Let  a1  be  the  slope  at  O,  and  dl  the  deflection  at  the 
centre.     Then, 

i  wr  i  wr 

from  (B),  tan  a1  =  ->  -=rjr ;  and  from  (C),  ^  =  -5  -py . 
ID  &i  48  /j/ 

a.  B  A  EX*    5'    The    £irder    OA    rests 

-^Jj,    |          ^'       x  upon  supports  at  <9,  ^4,  and  carries 

P~~~C"  a  uniformly  distributed  load  of  in- 

tensity w. 

At   any  point  P  (x,  y)    of   the 
FlG-  318.  neutral  axis, 


Integrating, 

PTdy  __wl  a      wx* 

~fcc''^~4~X       ~~6~  T~CI> 

\  being  a  constant  of  integration. 

But  -j-  =  o  when  x  =  -,  and  therefore  c.  — . 

dx  24 

Hence, 


dx    "  4  6         24  ' 

Integrating, 

wl    ,       wx*      wl* 

—  Ely     =  — x" x  4-  cmt 

12  24  24 

:,  being  a  constant  of  integration. 

But  7  =  0  when  x  =  o,  and  therefore  c^  =  o. 
Hence 

—  £7?  =  —  ^r3  -  ^-  —  —  x.  .  (C) 

12  24  24  V    ' 


NEUTRAL   AXIS   OF  A    LOADED  BEAM.  441 

Let  ora  be  the  slope  at  O,  and  d^  the  deflection  at  the  centre. 
Then, 

from  (B),  tan  a,  =  —  ^  ;  and  from  (C),  <  =  ^  ~. 

Ex.  6.  A  girder  rests  upon  two  supports  and  carries  a  uni- 
formly distributed  load  of  intensity  w,  together  with  a  single 
weight  W  at  the  centre.  This  is  merely  a  combination  of 
Examples  4  and  5,  and  the  resulting  equations  are  : 


dx   '  ~  4  16  4  6 

and 


(B) 


12  ID  12  24         24 

Also,  if  A   is  the  slope  at  the  origin,  and  D  the  central 
deflection,  we  have,  from  (B), 

i  fwr    wr\ 

tan  A  =  £j(-rf-  +  —  J  =  tan  «,+  tan  «a ; 
and  from  (C), 


*=        +-•  -<+<• 


The  slope  and  deflection  of  an  arbitrarily  loaded 
girder  resting  upon  two  supports  may  be  determined  in  the 
manner  described  in  Art.  2. 

Let  C  be  the  lowest  point  of  the  deflection  curve.  The 
tangents  at  C  and  O  will  intersect  in  a  point  T  which  is  ver- 
tically below  the  centre  of  gravity  of  the  bending-moment 
area  corresponding  to  OC. 

Denote  this  area  by  Fand  the  horizontal  distance  of  centre 


442  THEORY  OF  STRUCTURES. 

of  gravity  from  OY  by  x.     Let  OL  be  the  angle  between  OT 
and  CT  produced.     Then 

F  d 


d  being  the  maximum  deflection. 

In  Ex.  6,  e.g.,  the  girder  being  symmetrically  loaded, 

i  wi  i     2  wr  i  wr  2  i    wr  5  / 

F= sr-     and     Fx  =  —^ 4- — ~-  —  EId 

242^382  1632        24  82 

Ex.  7.  Suppose  that  the  end   O  of  the  girder  in   Ex.   5 

R  R^        is  fixed.     The   fixture  introduces 

\j(^\\  a  left-handed  couple  at  0 ;  let  its 

°K---.,_al ~~~^k *  moment  be  M,. 

P  Let  the  reactions  at  O  and  A 

y  be  Rt ,  R^ ,  respectively. 

At  any  point  P  (x,  y]  of  the 
neutral  axis, 


But  M,  i.e.,  —  £7-r4,  is  zero  when  x  =  /. 


™-^  ......     (2) 


Integrating  eq.  (i), 


There  is  no  constant  of  integration,  as  -,-  =  o  when  x  =  o. 

dx 

Integrating  eq.  (3), 

-  Ely  =  R?*^  -  M?-  .  (4) 

1  6        24  J  2 

There  is  no   constant   of  integration,  as  x  and  y  vanish  to- 
gether. 


NEUTRAL   AXIS   OF  A    LOADED  BEAM. 

But  y  also  vanishes  when  x  =  /,  so  that 


443 


Hence,  by  eqs.  (2)  and  (5), 


and  so     tf,  =     «//.      .     (6) 


Thus,  the   b  ending-moment,  slope,  and   deflection  equations 
are,  respectively, 


(*> 


=<"*•  -•- 


.  I.  The  bending  moment  is  nil  at  points  given  by 

5  w  ,       wl* 

^f=  o  =  -golx  —  —x*  --  g-, 


F 


i.e.,  when  x  =  —  or  /.     Take  OF=  — . 
4  4 

Since  -^-7  =  o,  .F  is  a  point  of  inflexion. 

If  the  girder  is  cut  through  at  this  point, 

and    a    hinge    introduced    sufficiently 

strong  to  transmit   the  shear  (=  %wl),  »FIG.  320. 

the  stability  of  the  girder  will  not  be  impaired. 

Hence,  the  girder  may  be  considered  as  made  up  of  two 
independent  portions,  viz. : 

(a)  A  cantilever  OF-oi  length  — ,  carrying  a  uniformly  dis- 
tributed load  of  intensity  w,  together  with  a  weight  £  wl  at  F. 


444  THEORY  OF  STRUCTURES. 

The  maximum  bending  moment  on   OF  is  at  O,  and  is 
3      f  /   ,   wl  I      wr 
=  8W/4+78=:l~' 

(b)  A  girder  FA  of  length  —  ,  carrying  a  uniformly  distrib- 

uted load  of  intensity  w. 

The  maximum,  bending  moment  on  FA  is  at  the  middle 


This  result  may  also  be  obtained  from  eq.  (7)  by  putting 

—  --  —  o.     Whence 
dx 

O  =  %wl  —  wx,     or     x  =  f/, 
.and  therefore 


The  shearing  force  and  bending  moment  at  different  points 
•of  the  girder  may  be  represented  graphically  as  follows  : 

The  shearing  force  at  any  point  of  which  the  abscissa  is  x  is 

S  =  f  wl  —  wx. 

Take  OB  and  A  C,  respectively  equal  or  proportional  to  •§•  wl 
and  fw/;  join  BC.  The  line  BC  cuts  OA  in  /?,  where  OP  =  f/. 
The  shearing  force  at  any  point  is  represented  by  the  ordinate 
between  that  point  and  the  line  BC. 

The  bending  moment  at  the  point  (x,  y)  is 


Take  OG,  DE,  and  OF,  respectively  equal  or  proportional 

wr      g  I 

to  -5—,  --^w/2,  and  -.     The  bending  moment  at  any  point  is 
o        I2o  4 

represented  by  the  ordinate  between  that  point  and  the  parab- 
ola passing  through  G,  F,  and  A,  having  its  vertex  at  E  and 
its  axis  vertical. 


NEUTRAL  AXIS  OF  A   LOADED  BEAM.  445 

dy 

Cor.  2.  The  deflection  is  a  maximum  when  —  =  o,  i.e.,  when 


or  at  the  point  given  by  x  =  -g05  —  1/33). 

Substituting  this  value  of  x  in  eq.  (9),  the  corresponding 
value  of  jj/  may  be  obtained. 

Ex.  8.  If  both  ends  of  the  girder  in  eq.  (7)  are  fixed,  the 

wl 
reaction  at  each  support  is  evidently  — ,  and  the  equation  of 

moments  becomes 

T^jd^y  _  wl         wx* 

Integrating, 

dy       wl          w 

—  El-f-  =  — x*  —  £**  —  M.x (2) 

dx        \  6 

dy 
No  constant  of  integration  is  required,  as  -r-  =  o  when  x  =  o. 

-j-  is  also  zero  when  x  =  I  (also  when  x  =  — ). 
dx  V  21 


and  hence 

^'  =  1? (3) 

Integrating  eq.  (2), 


12  24  24 

There  is  no  constant  of  integration,  as  x  and  y  vanish  together. 
The  central  deflection  I  i.e.,  when  #  =  —)=*  ~^~~cj' 


44-6  THEORY  OF  STRUCTURES. 

If  the  load,  instead  of  being  uniformly  distributed,  is  a 
weight  ^concentrated  at  the  centre,  then,  for  one  half  oi  the 
girder, 


Integrating, 

dv       W 
—  £/•-/-  =  —x*  -  M,x  ......     (6) 

dx        4 

dy 
There  is  no  constant  of  integration,  as  -j-  =  o  when  x  =  o. 

dy  I 

-j-  is  also  evidently  zero  when  x  =  —  ,  and  hence 

W  I  Wl 

0  =  _/>_^-,     or     M$~-g-.      •   V   .     (7) 

Integrating  eq.  (6), 

W  x*       W  Wl 

-Efy=—x*-Ml-  =  —x*-~x\      ...     (8) 

12  '2          12  l6 

There  is  no  constant  of  integration,  as  x  and  y  vanish  together. 

I     Wl 

The  central  deflection  =  ---  p-=-. 

192  El 

4.  Supports  not  in  same   Horizontal   Plane.  —  In   the 

preceding  examples  it  has  been  assumed  that  the  ends  of  the 
girder  are  in  the  same  horizontal  plane.  Suppose  that  one  end, 
e.g.,  A,  falls  below  O  by  an  amount  jj>,  ,  yl  being  small  as  com- 
pared with  /. 

The  abscissae  of  points  in  the  neutral  axis  are  not  sensibly 
changed,  but  the  conditions  of  integration  are  altered.  Con- 
sider Ex.  4. 

Between  O  and  C, 

W 


NEUTRAL   AXIS  OF  A    LOADED   BEAM.  447 

Integrating, 


cl  being  a  constant  of  integration. 
Integrating  again, 


W 
-  Ely  =  —*>  +  c>*  ......     (3) 


There  is  no  constant  due  to  the  last  integration,  as  x  and  y 
vanish  together. 
Between  C  and  A, 

W  f          A        W 

"*- 

Integrating  twice, 

-EIdj-x  =  -~(l-xf  +  c,,      ......    (5) 

and 

W 
-Efy=—(l-*r  +  cs  +  c..     .....     (6) 

c^  ,  <:,  being  constants  of  integration. 

The  tangent  at  C  is  no  longer  horizontal,  but  makes  a  defi- 

dy 
nite  angle  6  with  the  horizon,  so  that  -j-  is  now  tan  6  when 

x  —  —  .    Also,  the  values  of  -=-  and  y  at  C,  viz.,  tan  0  and  d,  as 

given  by  eqs.  (2)  and  (3),  must  be  identical  with  those  given  by 
eqs.  (5)  and  (6),  while  the  value  y,  at  A,  as  given  by  eq.  (6) 
when  x  =  /,  is  equal  to  yl  .  Therefore 

W  W 

—?  +  ^  =  -  EI  tan  0  =  -  ^t'  +  c,, 


r?  +  Ci      =  _  EId  =        p  +  C       +  c 
90  '2  96  32 

and 


44^  THEORY  OF  STRUCTURES. 

Hence, 

v        W  v       W  W 

' 


fully  defining  both  halves  of  the  neutral  axis. 

dy 
Again,  in  Ex.   6  it   is  no   longer  true  that  -^-  =  o  when 

x  =  — ,  but  the  conditions  of  integration  are  y  =  o  when  x  =  o, 

dy 

and  y  =  yl  when  x  =  /.     These,   together  with  -y-  =  o  when 

dx 

x  =  o,  are  also  the  conditions  in  Ex.  7.     Other  cases  may  be 
similarly  treated. 

5.  To  Discuss  the  Form  assumed  by  the  Neutral  Axis 
of  a  Girder  OA  which  rests  upon  Supports  at  O  and  A, 
and  carries  a  Weight  JP  at  a  Point  B,  distant  r  from  O. 

Let  OBA  be  the  neutral  axis  of  the  deflected  girder. 

The  reactions  at  O  and  A  are 


c . ,      , 


x       /  -  r 


P — -j —  and  P-j,  respectively. 
1  /  / 

Let  BC,    the   deflection   at   Ct 


Let  a  be  the  slope  of  the  neutral  axis  at  B. 
The  portions  OB,  BA   must  be  treated  separately,  as  the 
weight  at  B  causes  discontinuity  in  the  equation  of  moments. 
First,  at  any  point  (x,  y)  of  OB, 


Integrating, 


cl  being  a  constant  of  integration. 


NEUTRAL  AXIS  OF  A   LOADED  BEAM.  449 

dy 
But  -r-  =  tan  a  when  x  =  r,  and  therefore 


/  —  rr* 
-El  tan  a  =  P—  —  j  +  *,  . 


Hence, 
Integrating, 


Ti-T-'v-  •  •  (3) 


There  is  no  constant  of  integration,  as  x  and  y  vanish  to- 
gether. 

Also,  y  =  d  when  x  =  r. 


-  r  tan  a)  =  -  P--.    ...     (4) 

In  the  same  manner,  if  A  is  taken  as  the  origin,  and  AB 
treated  as  above,  equations  similar  to  (i),  (2),  (3),  and  (4) 
will  be  obtained,  and  may  be  at  once  written  down  by  sub- 

stituting in  these  equations  n  —  a  for  a,  PT-  for  P^^-,  l—r 

I  I 

for  r,  and  r  for  /  —  r. 

Thus,  the  equation  corresponding  to  (4)  is 

-  El  {  d  -  (I  -  r)  tan  (*  -  a)  }  =  -  P  '-  iLU^u       (5) 

o 

Subtracting  (5)  from  (4), 

p 


na==-r-r-2r);   ....     (6) 
and  from  (4), 


450  THEORY   OF  STRUCTURES. 

Thus,  eqs.  (2)  and  (3)  become 


and 

-Ely     =  ?'-*>-?(l-r)(2l-r)*,.     .     (9) 


the  latter  being  the  equation  to  the  portion  OB  of  the  neutral 
axis,  and  the  former  giving  its  slope  at  any  point. 
Next,  at  any  point  (x,y)  of  BA, 


z  P— j-x  -  P(x  -  r) (10) 

Integrating, 


ca  being  a  constant  of  integration. 

dy 

But  -j-  =  tan  a  when  x  =  r. 
dx 

••  pl—r  ? + '• =  ~  EI  tan " =  ~  f  7(/  ~ r)(/  ~ 2r 

and 

Pl-r  ,  ,        N 
c9=--—-r(2t-r). 

Hence, 

-  £/£ = T-r1*'  -  ?(^  - r)i  -  S(/  - r)(2/  - r)- 

Integrating, 

-  Ely  =  g1-^^'  ~  g(*  -  ^  -  £ /('  - 

£4  being  a  constant  of  integration. 
But  y  —  d  when  x  =  r. 


NEUTRAL   AXIS   OF  A   LOADED  BEAM.  45  1 


and  £4  =  o.     Hence, 

-  Ely  =  >  -     (*  -  r)'  -Pr(l-r)(2/-r)x,      (12) 


which  is  the  equation  to  the  portion  BA  of  the  neutral  axis, 
eq.  (11)  giving  its  slope  at  any  point. 

In  the  figure  r  <  —  ,  and    the    maximum  deflection   of  the 

girder  will  evidently  lie  between  B  and  A,  at  a  point  given  by 

dy 
putting       =o  in  eq.  (12),  which  easily  reduces  to 


and  therefore 


X  =  I  — 


is  the  abscissa  of  the  most  deflected  point.     The  corresponding 
deflection  is  found  by  substituting  this  value  of  x  in  eq.  (12). 

If  r  >  —  ,  the  maximum  deflection  lies  between  O  and  B,  at 

dy 
a  point  determined  by  putting  -^-  =  o  in  eq.  (8),  which  then 

easily  reduces  to 


from  which 


fr(2l  -  r) 

"V"      .< 


452  THEORY  OF  STRUCTURES. 

Substituting  this  value  of  x  in  eq.  (9), 


the  maximum  deflection  =  —gj — j—  I )  . 

EXAMPLE.— P  =  15,000  Ibs.,  /  =  100  ft.,  r  =  go  ft. 
The  distance  of  most  deflected  point  from  O 


/go  X  1 10  , 

=  y  -  — —  =  57-44  ft., 


and  the  maximum  deflection 

_  15000       10  too  X  iioy      500000 
~  W  X  I5o  V       3       /   =  ~~^T~(33)  * 


6.  To  Discuss  the  Form  of  the  Neutral  Axis  of  a  Girder 
OA  which  rests  upon  Supports  at  O  and  A  and  carries 
several  Weights  JP, ,  J*a ,  1% ,  .  .  . ,  at  points  i,  2,  3,  .  .  .  ,  of 
which  the  Distances  from  O  are  r0  ra ,  r, , .  . . ,  respectively. 


A-X 


FIG.  322. 

It  may  be  assumed  that  the  total  effect  of  all  the  weights 
is  the  sum  of  the  effects  of  the  separate  weights,  and  thus  each 
may  be  treated  independently,  as  in  the  preceding  article. 

Let  al ,  ara ,  or3 ,  .  . .  be  the  slopes  at  the  points  i,  2,  3,  .  .  . 
of  the  neutral  axis. 

Considering  Pt ,  the  equation  to  Oi  is 

-  Ely  =      '-V  -        '(/  - 


NEUTRAL  AXIS  OF  A    GIRDER   OA.  453 

and  to  lA, 

-  Ely  =  §^V  -  §(*  -  r,)'  -  §  T-j(l-  r^(2l-  r>. 
Considering  P,  ,  the  equation  to  O2  is 

-  Ely  =  §  ~V  -  §  ^/  - 


and  to  2A, 


l 


-6    T    ~        -  •  -      /    - 

and  so  on  for  Pz  ,  P4  ,  etc. 

The  total  deflection  F  at  any  point  (x,  F)  is  the  sum  of  the 
deflections  due  to  the  several  loads. 

Take,  e.g.,  a  point  between  3  and  4,  and  let  dl  ,  d^  ,  d^  ,  .  .  . 
be  the  deflections  of  this  point,  due  to  Plt  P9,  Pa  ,  .  .  .  ,  respec- 
tively. Then 


-Eld,  = 


-EU  = 


and  so  on.     Hence, 

- 


(A) 


454  THEORY  OF  STRUCTURES. 

Again,  the  position  of  the  most  deflected  point  is  found  by 
making  -7-  =  o  in  the  equation  to  that  portion  of  the  neutral 

axis  between  two  of  the  weights  in  which  the  said  point  lies. 
The  result  is  a  quadratic  equation,  and  the  value  of  x  derived 
therefrom  may  be  substituted  in  eq.  (A),  which  then  gives  the 
maximum  deflection. 

EXAMPLE.—  A  girder  of  100  ft.  span  supports  two  weights 
of  20,000  Ibs.  and  30,000  Ibs.  at  points  distant  respectively  20  ft. 
and  60  ft.  from  one  end. 

The  most  deflected  point  must  evidently  lie  between  the 
two  weights,  and  the  equation  to  the  corresponding  portion  of 
the  neutral  axis  is 


x*  2OOOO 

—  EIY  =  ^(20000  X  80  +  30000  X  40)  —  —  g—  (*  —  2°)3 

-  -^-(20000  X  20  X  80  X  1  80  +  30000  X  60  X  40  X  140) 

14000  ,   i  oooo. 

=  —  -  x  —  -  (x  —  20)  —  26400000^. 

Kis  a  maximum  when 

dY 

—  -  —  o  —  14000^  —  ioooo(^  —  20)  —  26400000, 

or 

X*  +  lOO^r  —-7600  =  O, 

or 

x  =  50.497  ft. 

Remark.  —  Instead  of  assuming  -jr  =  ^cEI-j-^  ,  it  would  be 

77  T  JA 

more  accurate  to  take  -jz-  =  ±  El  cos  6  —  (Art.  2),   and  the 

first  integration  would  make  the  left-hand  side  of  the  slope 
equation  ±  El  sin  0  instead  of  ±  El  tan  6. 


MOMENT  OF  INERTIA  VARIABLE. 


455 


7.  Moment  of  Inertia  variable.  —  In  the  preceding  exam- 
ples the  moment  of  inertia  /  has  been  assumed  to  be  constant. 
From  the  general  equations, 


-->.    j     2   -        , 

dx       c 

c  being  proportional  to  the  depth  of  the  girder  at  a  transverse 
section  distant  x  from  the  origin. 

Hence,  for  beams  of  uniform  strength,  the  value  of  c  in 
terms  of  x  may  be  substituted  in  the  last  equation,  which  may 
then  be  integrated. 

Again,  let  Fig.  323  represent  a  cantilever  of  length  /,  spe- 
cific weight   w,    circular  section, 
and  with  a  parabolic  profile,  the 
vertex  of  the  parabola  being  at  A. 

Let  2b  be  the  depth  of  the 
cantilever  at  the  fixed  end. 

Let  the  cantilever  also  carry  a 
uniformly  distributed  load  of  in- 
tensity/. 

Consider  a  transverse  section  of  radius  z  at  a  distance  x 
from  the  fixed  end. 

Let  x,  y  be  the  co-ordinates  of  the  neutral  axis  at  the  same 
section.  Then 


But  *'  =  -/_ 


FIG.  323. 


or 


Integrating, 


4~  r 


wnV 
~6~7~ 


-7    +  *T-       •     -     (0 


456 


THEORY  OF  STRUCTURES. 


dy 
There   is  no  constant  of  integration,  as  -—  =  o  when  x  =  o. 

Integrating  again, 

px* 


There  is  no  constant  of  integration,  as  x  and  jj/  vanish  to- 
gether. Thus,  equation  (i)  gives  the  slope  at  any  point,  and 
equation  (2)  defines  the  neutral  axis. 


The  slope  at  the  free  end  (*  =  /)  =  -m      + 


The  deflection  •« 


~& 


8.  Springs  Fixed  at  One  End  and  Loaded  at  the  other 
with  a  Weight  W. 

Data.  —  Length  =  /;  breadth  =  b,  and  depth  =  d  at  fixed 
support  ;  V=  volume  of  spring;  f=  maximum  coefficient  of 
strength  ;  A  =  maximum  deflection. 

CASE  a.  Simple  rectangular  spring. 

By  Ex.  i,  Art.  39, 


_ 
-"' 


since 


Wl  _M  _2f  _  1207 

' 


Also, 


y  2// 


61    ' 


Hence, 


The  work  done  =  —  = 


(3) 


SPXJNGS. 


457 


CASE  b.  Spring  of  constant  depth  but  triangular  in  plan. 
Let  bx  be  the  breadth  at  a  distance  x  from  the  fixed  end. 
Then 

b        I  -  x 


and  /  at  the  same  point 


l-X 


12 


12 


bd\ 


FIG.  325. 


1207 

EbcT' 


Integrating  twice, 


and 
Also, 

dy  _ 
dx~ 

12WI 

(A\ 

6WI  , 

•V 

y 

•   A 

E$3r*  ' 

6wr    /r 

T/T7  A 

Ebd*  "Ed  
bd*ffP      f*bdl      f*V 

(5) 

Hence, 
/.  V  — 

The  work  done  = 

61  Ed       6E      '  $£' 

r  

WA      f*V 

(6) 

N.B. — The  results  I  to  6  are  the  same  if  the  springs  are 
compound  ;  i.e.,  if  the  rectangular  spring  is  composed  of  n 
simple  rectangular  springs  laid  one  above  the  other,  and  if  the 
triangular  spring  is  composed  of  n  triangular  springs  laid  one 
above  the  other. 


458  THEORY   OF  STRUCTURES. 

CASE  c.   Spring  of  constant  width  but  parabolic  in  elevation. 
Let  dx  be  the  depth  at  a  distance  x  from  the  fixed  end. 
Then 


dl -     / 

and  /at  the  same  point  = 


FIG.  326. 


bd:    wti-*\i 

12    '       12  I      /      /' 


d*y         W(J         \       l2Wl*f,         x 
,.^r=  _(/_*)=__.(/_,) 


Integrating  twice, 

^  =  T  ^  ^  *(/  ~  *}l~ 2/t(/ 

and  hence 


*KJL.    4//a 

E    bd*~ ->K<1 v) 


Also, 


The  work  done    = =-2 


2      ~6   £  ' 

9.  Girder  Encastre  at  the  Ends.— The  girder  BCDEFG 
rests  upon  supports  at  the  ends,  is  held  in  position  by  blocks 
forced  between  the  ends  and  the  abutments,  and  carries  a  uni- 
formly distributed  load  of  intensity  w. 


GIRDER   ENCASTRE   AT   THE   ENDS. 


459 


It  is  required  to  determine  the  pressure  that  must  be  devel- 
oped between  the  blocks  and  the  girder  so  that  the  straight 
portion  between  vertical  sections  at  points  O  and  A  of  the 


W.I 


FIG.  327. 

neutral  axis  may  be  in  the  same  condition  as  if  the  girder  were 
fixed  at  these  sections. 

Let  /  be  the  length  of  OA. 

Let  R  be  the  reaction  at  the  surface  BC>  and  r  its  distance 
from  O. 

Let  /fbe  the  reaction  between  the  block  and  the  end  CD, 
and  h  its  distance  from  O. 

Let  P  be  the  weight  of  the  segment  on  the  left  of  the  ver- 
tical section  O,  and  /  its  distance  from  O. 

Then  for  the  equilibrium  of  the  segment  on  the  left  of  the 
section  at  O, 


2  *  12 

•    /?  —  P        — 

.\K-:*>*~    2  , 


and 


I         wl\  wl* 

(P— — \r-Pp-- 
\  21  '  12 


—  the  required  pressure. 


Again,  take  O  as  the  origin,  OA  as  the  axis  of  x,  and  a 
vertical  through  O  as  the  axis  of  _y. 

At  any  point  (x,  ^/)  of  the  neutral  axis, 


_  E         =       x  _         _        .        (see  Ex.  8.) 

dx*  2  2  12  ^ 


460  THEORY  OF  STRUCTURES. 

10.    On   the  Work    done   in   bending   a    Beam. — Let 
A'B'C'D'   be   an    originally   rectangular    ele- 
\q'   ment  of  a  beam  strained  under  the  action  of 
jfg.    external  forces. 

Let  the  surfaces  AD',  B'C'  meet  in  G>; 
O  is  the  centre  of  curvature  of  the  arc  P  Q  of 
the  neutral  axis. 

Let  OP  =  R  =  OQf. 
Let  the  length  of  the  arc  P  Q'  =  dx. 
Consider  any  elementary  fibre/'/,  of  length 
dx'y  of  sectional  area  a,  and  distant  y  from  the 
neutral  axis. 

Let  t  be  the  stress  in  p'q' . 

The  work  done  in  stretching/'/ 


But     ---     and 


-jr--4 

dx9  -dx  y 


The  work  done  in  stretching/'/  =  -  — 

and  the  work  done  in  deforming  the  prism  A'B'C'D' 
iE 


Hence,  the  total  work  between  two  sections  of  abscissae 

-^a> 

C'-'iEI  ,       El  r'"dx 
=        .T.*-  I?" 


I      J/ 

But  -    —  —\  therefore  the  work  between  the  given  limits 


TRANSVERSE    VIBRATIONS.  461 

This  expression  is  necessarily  equal  to  the  work  of  the  ex- 
ternal forces  between  the  same  limits,  and  is  also  the  semi  vis- 
viva  acquired  by  the  beam  in  changing  from  its  natural  state 
of  equilibrium. 

Cor. — If  the  proof  load  P  is  concentrated  at  one  point  of  a 

p 
beam,  and  if  d  is  the  proof  deflection,  the  resilience  =  —d. 

If  a  proof  load  of  intensity  w  is  uniformly  distributed  over 
the  beam,  and  if  y  is  the  deflection  at  any  point,  the  resili- 
ence =  —  /  wydx,  the  integration  extending  throughout  the 

whole  length  of  the  beam. 

The  case  of  the  single  weight,  however,  is  the  most  useful 
in  practice. 

ii.  On  the  Transverse  Vibrations  of  a  Beam  resting 
upon  Two  Supports  in  the  same  Horizontal  Plane. 

It  is  assumed — 

(a)  That  the  beam  is  homogeneous  and  of  uniform  sectional 
area. 

(b)  That  the  axis  (neutral)  remains  unaltered  in  length. 

(c)  That  the  vibrations  are  small. 

(d)  That  the  particles  of  the  beam  vibrate  in  the  vertical 
planes  in  which  they  are  primarily  situated.     In  reality,  these 
particles  have  a  slight  angular  motion  about  the  horizontal  axis 
through  the  centre  of  gravity  of  the  section,  but  for  the  sake 
of  simplicity  the  effect  of  this  motion  is  disregarded. 

Jif  M44M 


/•I  A 


dx 



A-X 

Y 

CiYC' 
i  S+dS 
w  dx 

FIG.  329. 

Let  OA  be  the  beam. 

Take  O  as  the  origin,  the  neutral  line  OA  as  the  axis  of  x, 
and  the  vertical  O  Fas  the  axis  of  y. 

Consider  an  element  of  the  beam,  bounded  by  the  vertical 


4^2  THEORY  OF  STRUCTURES. 

planes  BC,  B ' C ',  of  which  the  abscissae  are  x  and  x  -f-  dx, 
respectively. 

Let  w  be  the  intensity  of  the  load  per  unit  of  length ;  hence 
wdx  is  the  load  upon  the  given  element,  and  acts  vertically 
through  its  centre. 

Let  5  be  the  shearing  force  at  B ;  5  -f-  dS  the  shearing 
force  at  B'. 

Let  M  be  the  bending  moment  at  B]  M  -\- dM  the  bend- 
ing moment  at  B'. 

iv  d*  v 

Also,  the  resistance  of  the  element  to  acceleration  = —-. 

g  at 

Hence,  at  any  time  t, 


w      d  y 

—dx  -j-f  +  5  —  (S  -f-  dS)  —  wdx  =  o, 


or 


_ 

d?       w  dx 


-<r  =  o (O 


Again,  taking  moments  about  the  middle  point  of  BBf  or 
CO, 

dx 

T  =  0' 


or 


dx 
But  M  =  —  EI--^, .     Therefore 


dM 

--T-  =  S. (2) 

^  ' 


vT          j  ds       KT 

—EI-^-.,     and     -r-  =  —  EI~. 
dx* '  dx  dx* 


Hence,  from  (i), 


CONTINUOUS   GIRDERS.  463 

This  equation  does  not  admit  of  a  finite  integration,  but 
may  be  integrated  in  the  form  of  a  partial  differential  equation. 

12.  Continuous   Girders. — When  a  girder  overhangs  its 
bearings,  or  is  supported  at  more  than  two  points,  it  assumes  a 
wavy  form  and  is  said  to  be  continuous.     The  convex  portions 
are  in  the  same  condition  as  a  loaded  girder  resting  upon  a 
single  support,  the  upper  layers  of  the  girder  being  extended 
and  the  lower  compressed.     The  concave  portions  are  in  the 
same  condition  as  a  loaded  girder  supported  at  two  points,  the 
upper  layers  being  compressed   and  the  lower  extended.     At 
certain  points,  called  points  of  contrary  flexure,  or  points  of  in- 
flexion, the  curvature  changes  sign  and  the  flange  stresses  are 
necessarily  zero.     Hence,  apart  from  other  practical1  considera- 
tions, the  flanges  might  be  wholly  severed  at  these  points  with- 
out endangering  the  stability  of  the  girder. 

13.  The  Theorem  of  Three  Moments. — It  is  required  to 
determine  a  relation  between  the  bending  moments  at  any  three 
consecutive  points  of  support  of  a  loaded  continuous  girder  of 
several  spans. 


Rr-1                       Rr 

r               ! 
A           ix 

Rr+1 

!v 

1     -3J 

V 

FIG.  330. 

Let  O,  X,  V  be  the  (r  —  i)th,  rth,  and  (r  +  l)th  supports, 
respectively. 

Let  OX=tr,XV=tr  +  l. 

CASE  A.  Let  wr  be  the  load  per  unit  of  length  on  OX, 
wr  +  1  the  load  per  unit  of  length  on  XV. 

Let  Rr.i,  Rr,  ftr  +  i  be  the  reactions  at  O,X,  V,  respectively. 

Let  Mr^,  Mr,  Mr+l  be  the  bending  moments  at  O,  X,  V, 
respectively. 

Let  a  be  the  angle  which  the  tangent  to  the  girder  at  X 
makes  with  OV. 

Consider  the  segment  OX,  and  refer  it  to  the  rectangular 
axes  Ox,  Oy. 


464  THEORY  OF  STRUCTURES. 

The  equation  of  moments  at  any  point  (x,  y)  is 

Rr.lX-wr~  +  Mr.,  =  M.      .    .     (i) 


lr 

.;Rr_ltr-wr--  +  Mr_l  =  Mr (2) 

Similarly,  the  segment  XV  gives 

Combining  (2)  and  (3), 


r_  r1Ir  r 

=  J/X4 +  /,  +  ,) (4) 

Integrating  (i), 

-£'%  =  Xr-^-*>r%  +  Jfr.S  +  C,         •       •       (5) 

c  being  a  constant  of  integration. 

dy 
When  x  —  lry  -j-  =  tan  a. 

.-.  -  El  tan  a  =  Rr_,  |-  -  «v|r  +  ^.^  +  c,       .     .     (6) 
Integrating  (5), 

-  Ely  =  Rr-^  -  v>r  ~  +  Mr_t  ~  +  ex.      .     (7) 

There  is  no  constant  of  integration,  as  x  and  y  vanish  together. 
Also,  y  =  o  when  x  —  lr . 


or 

c  =  —  Rr  1 4-  -\-  v>r  —  —  Mr  ,  -.  (8) 

r  —  izr  r   4»  ^  r  — 1/«  ..          \/ 


THE    THEOREM  OF    THREE  MOMENTS.  465 

Substituting  this  value  of  c  in  eq.  (6), 

-  EI  tan  a  =  Rr.,  ^  -  wr£  +  Jfr.t  j  .....     (9) 
Similarly,  the  segment  XV  gives 


Adding  eqs.  (9)  and  (10),  transposing,  and  simplifying, 

Rr-Jr    +  Rr  +  fr  +  i 

=  WS  +  f«V  +  ,/',+  .  -  \Mr_Jr  -  PVr+I.    (ii) 
Finally,  combining  eqs.  (4)  and  (u), 

+  wr+I  rr+l).  (12) 


If  the  girder  is  supported  at  «  points,  there  are  w  —  2  equa- 
tions connecting  the  corresponding  bending  moments,  and  two 
additional  equations  result  from  the  conditions  of  support  at 
the  ends.  For  example,  if  the  ends  merely  rest  upon  the  sup- 

dy 
ports  Ml  —  o  and  Mn  =  o  ;  if  an  end  is  fixed,  -=-  =  o  at  that 

point. 

The  point  of  maximum  bending  moment,  the  points  of 
inflexion,  and  the  point  of  maximum  deflection  in  any  span  are 

dM  dv 

found  by  making  —=-  =  o,  M  —  o,  and  -j-  —  o,  respectively. 

Thus,  for  the  span  OX, 

dM 

—  —  o  =  Rr^  —  WyX  ; 

/.  x  =  —  —  f,    and  maximum  B.M.  =  --  —  -f-  Mr.t  \ 

"Wr  2      Wr 


a  quadratic  giving  x ; 


466  THEORY  OF  STRUCTURES. 


c, 


a  cubic  from  which  x  may  be  found  by  trial.  The  maximum 
deflection  is  obtained  by  substituting  the  value  of  x  in  eq.  (7) ; 
c  being  given  by  eq.  (8). 

CASE  B.  Let  the  loads  upon  OX,  XV,  respectively,  consist  oi 
a  number  of  weights  Pl ,  P9 ,  P3 ,  .  .  . ,  distant  pl , /2  ,/3 ,  .  .  .  from 
O,  and  Ql ,  <23 ,  <2, ,  .  .  . ,  distant  #,,£,,&,...  from  F.  Refer 
the  neutral  axis  OA X  to  the  rectangular  axes  Ox,  Oy. 

It  may  be  assumed  that  the  total  effect  of  all  the  weights 
is  the  algebraic  sum  of  the  effects  of  the  weights  taken  sepa- 
rately. 

Consider  the  effect  of  Pt  at  A. 

The  equation  of  moments  at  any  point  (x,  y)  of  the  neutral 
axis  between  O  and  A  is 


Integrating, 

cl  being  a  constant  of  integration. 
Integrating  again, 

-Efy^Rr-^+Mr-^  +  cs.      .     .     (3) 

There  is  no  constant  of  integration,  as*  and  y  vanish  together. 
The  equation  of  moments  at  any  point  (x,  y)  between  A  and 


(4) 


THE    THEOREM    OF   THREE   MOMENTS. 
Integrating, 

-El4jL  =  Rr.£-^(x-pl)'+Mr-*  +  ct.    ...     (5) 

Integrating  again, 

-  Ely  =  Rr.~  -     (*  -  A)3  +  Mr_      +  v  +  ,,  .       (6) 


dy 

Now,  at  the  point  A,  the  values  of  -7-  and  jj/,  as  given  by 

m& 

eqs.  (5)  and  (6),  are  identical  with  those  given  by  eqs.  (2)  and 
(3)  ;  also,  in  equation  (6),  y  =  o  when  x  =  lr. 
Hence, 


and 

o  =  R^  -  §(/r  -  A)3  +  ^r-~  +  cjr  +  c9  ; 

so  that 

£3  =  0, 

and 

«,  =  *.=  -^-,§  +  J(/,-/1)t-^-,j.  •    •    •    (7) 

Let  a  be  the  slope  at  X\  then,  by  eqs.  (5)  and  (7), 


-  EI  tan  a  =  R_       -       (/,  -  py(2lr  +/,)  +  Jfr_.   (8) 
Similarly,  the  segment  XV  gives 
-EI^(n-a}  =  Rj+Mr+      ......    (9) 


468  THEORY  OF  STRUCTURES. 

Adding  eqs.  (8)  and  (9),  and  transposing, 

*,_,/,'  +  JWV*  =  §&  -  A)'(2/r  +A) 

-f^.^-fJf^/^.      (10) 

Again,  taking  moments  about  Jf, 

Rr  jr  _  />('r-A)  +  -^-i  =  ^  =  ^WH-X  +  ^H-:>     00 
whence 
Rr  _,/;  +  ^H-x'V+x 


and  finally,  by  eqs.  (10)  and  (12), 

Mrjr  +  2Mr(lr  +  /,+1)  +  Mr+1tr+I  =  -  Pfa  -  A2)-     (13) 

The  effect  of  each  weight  may  be  discussed  in  the  same 
manner,  and  hence  the  relation  between  Mr_iy  Mr,  and  Mr+l 
may  be  expressed  in  the  form 


r_  jr  +  2Mr(lr  +  /r+1)  +  Mr+ltr+l  =  -  2fa  -  p  , 


Cor.  I.  The  relation  between  Mr_T,  Mr,  Mr  +  l  for  a  uni- 
formly distributed  load  maybe  easily  deduced  from  eq.  (14). 
For  example,  let  a  uniformly  distributed  load  of  intensity  wr 
cover  a  length  2a(<lr)  of  the  span  OX,  and  let  z  be  the  distance 
of  its  centre  from  O.  Then 


V  _/}  =  f  +^,(4<  -/)  =  ^;  -  *  - 

^r  *J  z-  a  r 


which  reduces  to  ~-f  when  z  =  a  =  •£. 


THE    THEOREM  OF   THREE  MOMENTS.  469 

Cor.    2.    Considering  the   rth   span  and    taking   moments 
about  the  rth  support, 


M  being  the  moment  of  the  load  on  the  span,  and  the  reaction, 
or  shear, 

M      Mr      Mr_, 


Hence,  the  shear  at  the  (r  —  i)th  support  for  the  rth  span 
=  the  reaction  at  the  same  support,  supposing  the  span 

an  independent  girder,  i.e.,  cut  at  its  supports, 
+  the  difference  of  the  forces,  or  reactions,  equivalent 

to  the  moments  at  the  supports. 

Again,  let  Mx  be  the  moment  of  the  load  on  the  segment 
x  with  respect  to  the  point  (x,y). 

Hence,  the  total  moment  about  (x,y) 


'M        ,r\  ,  Mr_,,,         .    ,  M. 


=  the  moment  at  the  same  point  supposing  the  span 

an  independent  girder. 

+  the  reactions  equivalent  to  the  moments  Mr_^ ,  Mr , 

multiplied  respectively  by  the  segments  lr  —  x  and  x. 

In  Fig.  331,  OX  being  the  rth  span,  let  OB X  be  the  curve 


FIG. 


of  bending  moments,  supposing  OX  an  independent  girder,  i.e., 
cut  at  O  and  X.  On  the  same  scale  as  this  curve  is  drawn, 
take  the  verticals  OE  and  XF  to  represent  Mr_^  and  Mr,  re- 


47°  THEORY  OF  STRUCTURES. 

spectively,  and  join  EF.     The  curve  OBX  corresponds  to  the 

(M  \ 

portion  \-j-x  —  Mx}  of  the  above  equation,  and  the  line  EF 

T 

Mr_J  \      Mr 

to  the  remainder,  i.e.,  —j—\lr—  xj  -f-  -j-x.  The  actual  bend- 
ing moment  at  any  point  of  OX  is  represented  by  the  algebraic 
sum  of  the  ordinates  of  the  curve  and  line  at  the  same  point, 
which  will  be  the  intercept  between  them,  since  they  represent 
bending  moments  of  opposite  kinds. 

Let  A  be  the  effective  moment  area,  or  the  algebraic  sum  of 
the  areas  for  the  load  and  for  the  moments  at  O  and  JT,  and 
let  x  be  the  horizontal  distance  of  its  centre  of  gravity 
from  O. 

Let  Ar  be  the  area  for  the  load,  i.e.,  the  area  of  the  curve 
OBX,  and  let  zr  be  the  horizontal  distance  of  its  centre  of 
gravity  from  O.  Then 

l'         ,       I, 

Ax  =  Arzr  +  Mr.=  +  -(Mr-  Mr_W 
=  Arzr  -f-  \Mr.j;  -f-  lMrlr\ 

This  result  will  be  referred  to  in  a  subsequent  article. 

14.  Applications. — EXAMPLE  i.  Swing-bridges  of  two 
spans  revolving  about  a  single  support  at  the  pivot  pier. 

This  is  a  case  of  a  girder  of  two  spans,  OX(=.  l^)y  XV(=  /2), 
resting  upon  supports  at  O  and  F,  and  continuous  over  a 
pier  at  X. 

The  bending  moments  at  O  and  Fare 
0  X  V    both  nil. 

T~~  T~  't          Let  M  be  the  bending  moment  at  X. 

FIG.  332. 

For  a  uniformly  distributed  load, 


or          =  - 


wt  being  the  intensity  of  the  load  on  OX,  w,  that  on  XV. 


SWING-BRIDGES.  47  l 

For  an  arbitrarily  distributed  load, 

l^  =  -A-B,    or    j|f=_I- 


where       A  =  2-(/*  -  /)     and     B  =  2(tJ  -  f). 

f,  *2 

Let  Rl  ,  ^2  ,  -/?3  be  the  reactions  at  O,  X,  V,  respectively. 
For  a  uniformly  distributed  load, 

P    -  w&    \M-  3w*/*>  ~ 

*~       "- 


2   "/-        8/^ 

i  ^- 

r- 


2 

For  an  arbitrarily  distributed  load, 

K        ^(A~/)   ,  M  _     P(lt-p)      I 

^•=  ~ 


R  =  2^_Zl}  +~=  ^Q(/*  ~  q)  -  -    A+B 

If  wl  =  O,  or  if  P  and  hence  ^4  —  o,  then  Rl  is  negative. 

So  if  wa  =  o,  or  if  Q  and  hence  B  —  o,  then  ^?3  is  nega- 
tive. 

Hence,  if  either  of  the  spans  is  unloaded,  the  reaction  at 
the  abutment  end  of  the  unloaded  span  is  negative  and  that 
end  is  subjected  to  a  hammering  action.  This  evil  may  be 
obviated  : 

(a)  By  loading  the  spans  sufficiently  to  make  R^  and  R^ 
zero  or  positive. 

This  result  is  attained  for  R, 


if  3Wl/,'  +  4W/.Y,  >  a/41,   or  « 
and  for  ^3 

if  4«vW  +  3^/,3  >  ^y,8,   or  if 


472  THEORY  OF  STRUCTURES. 

(b)  By  using  a  latching  apparatus  to  keep  the  ends  from 
rising. 

(c)  By  employing  suitable  machinery  to  exert  an  upward 
pressure,  at  least  equal  to  the  corresponding  negative  reaction 
upon  each  end,  which  is  thus  wholly  prevented  from  leaving  its 
seat. 

Cor.  i.    When  the  load   is  uniformly  distributed,  the  dis- 
tance x  of  the  point  of  inflection  in  OX  from  O  is  given  by 

M  =  o  =  R,x  --  —  ,  and  therefore   x  —  —  -  . 

2  W1 

Similarly,  the  distance  of  the  point  of  inflection  in  XV  from 
K=2A 

U\ 

If  /t  =  /,  =  /,  then 

M=-^l+w,  Rt  =  7-^^    *  =  = 

And  if  wl  =  2£/a  =  w,  then 


In  the  latter  case  —  -  =  £/  =  —  -3,  and  thus  a  hinge  may 

w1  w2 

be  introduced  in  each  span  at  a  distance  from  the  centre  pier 
equal  to  one  fourth  of  the  span,  without  impairing  the  stability 
of  the  girder.  Hence,  also,  the  continuous  girder  of  two  equal 
spans  may  be  considered  as  consisting  of  two  independent 
girders,  each  of  length  |/,  resting  upon  end  supports,  and  of 

two  cantilevers  each  of  length  —  . 

Ex.  2.  Swing-bridges  with  two  points  of  support  at  the 

'     t'        f 

h  {>  I* 


FIG.  333. 

pivot  pier,  as,  e.g.,  when  they  are  carried  upon  rollers  running 
in  a  circular  path. 


A  P  PLICA  TIONS.  47  3 

This  is  a  case  of  a  continuous  girder  of  three  spans. 

Let  I,  ,  /2  ,  /„  be  the  lengths  of  the  spaces,  w,  ,  «/,  ,  ze/3  the 
corresponding  intensities  of  the  loads,  which  are  assumed  to 
be  uniformly  distributed. 

Let  Rlt  R^j  R3,  Rt  be  the  reactions  at  the  supports  ; 
Ml  ,  Mt  ,  Ma  ,  Mt  the  corresponding  bending  moments.  Then 

-$(w1t:  +  wj;)',    .     (i) 
-}(«>X  +  wA).     .     (2) 

Let  the  ends  of  the  girders  rest  upon  the  supports,  and 
assume,  as  is  usually  the  case  in  practice,  that  the  centre  span 
is  unloaded,  i.e.,  that  w^  =  o.  Then 

Ml  =  o    and     M^  =  o. 
From  (i)  and  (2), 

A  =  -*»/,',   ••".   •   •   (3) 


and 

M,l,  +  2  !&,(!,  +  /,)=-&,!:  .....    (4) 
Hence, 

/./.V, 

' 


34" 
and 

_     ^/.34  -  W,V,  +  4) 


-  4(444 +  34' +  444 +  444)' 

Taking  moments  about  the  second  support, 


6/,V>  +  6/1VJ'  +  8/1'4/1)  +  Wj4V,_     _     (7) 


4(444  +  34'  +  444  +  444) 

Taking  moments  about  the  third  support, 


6/.V,  +  6/.V.'  +  8/,V,Q  +  w,n  (K. 

•     •     (°) 


4(444 +  34' +  444 +  444) 


474  THEORY  OF  STRUCTURES. 

Thus  Rt  and  jRt  are  both  positive  for  all  uniform  distribu- 
tions of  load  over  the  side  spans,  and  no  hammering  action 
can  take  place  at  the  ends. 

Again,  if  the  span  on  the  left  is  unloaded,  i.e.,  if  w1  =  o, 
M9  is  positive  and  M3  negative ;  and  if  the  span  on  the  right  is 
unloaded,  i.e.,  if  w3  =  o,  M,  is  negative  and  M3  positive. 

Thus,  at  the  piers,  the  flanges  of  the  girder  will  be  sub- 
jected to  stresses  which  are  alternately  tensile  and  compressive, 
and  must  be  designed  accordingly.  The  same  result  is  also 
true  for  arbitrarily  distributed  loads. 

Ex.  3.  The  weights  on  the  wheels  of  a  locomotive  passing 
over  a  continuous  girder  of  two  spans,  each  of  50  ft.,  taken  in 
order,  are  as  follows  :  15,000  Ibs.,  24,000  Ibs.,  24,000  Ibs.,  24,000 
Ibs.,  24,000  Ibs.  The  distances  of  the  wheels,  centre  to  centre, 
taken  in  the  same  order,  are  90  in.,  56  in.,  52  in.,  56  in.  Let 
it  be  required  to  place  the  wheels  in  such  a  position  as  to  give 
the  maximum  bending  moment  at  the  centre  pier. 

The  pier  must  evidently  lie  between  the  third  and  fourth 
wheels. 

Let  x  be  the  distance,  in  inches,  of  the  weight  of  15,000  Ibs. 
from  the  nearest  abutment.  The  remaining  two  weights  on 
the  span  are  respectively  x  -f-  90  and  x  -(-  146  in.  from  the 
same  abutment. 

The  two  weights  on  the  other  span  are  142—^  and  198— ;r 
in1.,  respectively,  from  the  nearest  abutment. 

Hence,  by  Case  B,  Art.  13,  if  M  is  the  bending  moment  at 
the  centre  pier, 


'  -  (x  +  90)' 


APPL1CA  TIONS. 


475 


Making  —  —  o  for  maximum  value  of  Mt  and  simplifying, 


dx 


and  therefore 


+  27648*  =  2518848, 


x  =  87.39  in.  =  7.28  +  ft. 


Thus,  the  B.  M.  at  the  centre  pier  is  a  maximum  when  the 
first  wheel  is  7.28  ft.  from  the  nearest  abutment. 

The  maximum  B.  M.  in  inch-pounds  is  obtained  by  substi- 
tuting x  =  87.39  in.  in  the  above  equation. 

15.  Maximum  Bending  Moments  at  the  Points  of  Sup- 
port of  Continuous  Girders  of  n  equal  Spans. 

Let  the  figure  represent  a  continuous  girder  of  n  spans,  I, 
2,  3,  ...»  —  I  being  the  n  —  I  intermediate  supports. 

o        123 r—  I     r     r-\-i »— I     n 

"A       A       A       A       A       A       A       A       A      A      A~ 


CASE  I.  Assume  all  the  spans  to  be  of  the  same  length  /, 
and  let  wl ,  wt ,  .  .  .  wn_l ,  wn  be  the  intensities  of  loads  uni- 
formly distributed  over  the  1st,  2d,  ...(»  —  i)th  and  nth  spans, 
respectively. 

By  the  Theorem  of  Three  Moments, 


-( 

4      l 

_      _/_' 

4      2 

_/_' 

4 

/2 
?8  +  4^4  +  ^6  =  —  -(w.  +  w6)  ; 


(O 

(2) 

(3) 
(4) 
(5) 


4/6  THEORY  OF  STRUCTURES. 

r 

mn_3  +  4mn_2  +  »/«_,  =  -  -(wn_2  +  wn.r)  ;        (n—  2) 

r 

mn.2  +  4w«_T  =  —  -<>«-i  +  Wn)-  (n  —  i) 

4 

?#0  and  ;#„  are  both  zero,  as  the  girder  is  supposed  to  be  rest- 
ing upon  the  abutments  at  o  and  n. 

From  these  (n  —  i)  equations,  the  bending  moments  ml  , 
m^  ,  .  .  .  mn_i  may  be  found  in  terms  of  the  distributed  loads. 

Eliminating  m^  from  2  and  3, 

r 

mi  _  x  5^3  _    4^4  =  _  -  j  (Wf  +  ^3)  -  4K  +  ^4)}-  •     •    (*0 
Eliminating  wg  from  4  and  x^  , 


Eliminating  MI  from  5  and 

m  — 


. 
Finally,  by  successively  eliminating  w6  ,  m6 

±     ««-!«««  -X 

/a 

--Uw»  +  w3)-4K  + 


the  upper  or  lower  sign  being  taken  for  the  terms  within  the 
brackets  according  as  n  is  odd  or  even,  and  the  coefficients 
^»-i  ,  #*-2  ,  tf«-3  ,  •  •  •  bein'g  given  by  the  law, 


APPLICATIONS.  477 

a,  =  4at  —  a,  =  209  ; 


#a  =  4*1        =    4  ; 
«,  =  I. 

Commencing  with  equations  n  —  3  and  n  —  2,  and  proceed- 
ing as  before, 

0«_x»*i   ±   *««-i 

/i 

=  —  -  \  an-9(u>i  +  «/,)  —  rf«_3(wa  +  ze;3)  +  d«_4(ws  +  «;«)  —  .  .  . 

±  1  5O«_4  +  w«_3)  =F  4(w«-3  +  «>„_,)  ±  (^«-2  +  «V_f)  }  ,     (^) 

the  upper  or  lower  sign  being  taken  for  the  terms  within  the 
brackets  according  as  n  is  odd  or  even. 
Solving  the  two  equations  y  and  2, 

r 

«,(*f«-x—i)=  -  -|^«-i^-2 


±  (<*n-i  +  a»-3—an-*Wn-i  =F 
and 

±  ^«-i(tfVi  -  i)  = 


/a  ( 

—  -  1  —  an_2w, 
4  » 


Hence,  since  w19wtt...wn  are  positive  integers,  the  value  of 
mH  will  be  greatest  when  «;,  ,  «/a  ,  w4  ,  we  ,  o;8  ,  .  .  .  are  greatest 
and  wa  ,  w5  ,  wn  ,  .  .  .  are  least  ;  and  the  value  of  mn_v  will  be 
greatest  when  ww  ,  wn^  ,  w«_3,  wM_4,  .  .  .  are  greatest,  and  wn_2  , 
wM-4  ,  WM_«,  .  .  .  are  least.  In  other  words,  the  bending  moments 
at  the  1st  and  (n  —  i)th  intermediate  supports  have  their  maxi- 


47  8  THEORY  OF  ^STRUCTURES. 

mum  values  when  the  two  spans  adjacent  to  the  support  in 
question,  and  then  every  alternate  span,  are  loaded,  and  the  re- 
maining spans  unloaded. 

#?,,#?,,...  mn-i  may  now  be  easily  determined. 

Thus,  by  eq.  (i), 

r 

m*  =  —  7^«  +  w^  ~  4wi 
4 

r  (  4 

=  -  -7  J  K  +  wj  -  -i  -  -a 

4     l  " 


a  +  .  .  .  I- 
But  an_,  =  4an_2  —  an_y 


and  is  greatest  when  w9,  w3,  wt,  w1f  .  .  .  are  greatest  and 
w4  ,  w,  ,  ws  ,  .  .  .  are  least. 

Similarly,  by  eqs.  (i)  and  (2), 


CONTINUOUS  GIRDERS.  479 

and  is  greatest  when  wl ,  wa ,  w^ ,  wt ,  w6 ,  .  .  .  are  greatest  and 
wa ,  w6 ,  w7 ,  w9 ,  .  .  .  are  least. 

Thus,  the  general  principle  may  be  enunciated,  that  "  in  a 
horizontal  continuous  girder  of  n  equal  spans,  with  its  ends 
resting  upon  two  abutments,  the  bending  moment  at  an  inter- 
mediate support  is  greatest  when  the  two  spans  adjacent  to 
such  support,  and  the  alternate  spans  counting  in  both  direc- 
tions, carry  uniforrnly  distributed  loads,  the  remainder  of  the 
spans  being  unloaded." 

CASE  II.  The  principle  deduced  in  Case  I  also  holds  true 
when  the  loads  are  distributed  in  any  arbitrary  manner. 

Consider  the  effect  of  a  weight  w  in  the  rth  span  concen- 
trated at  a  point  distant/  from  the  (r  —  i)th  support. 

By  the  Theorem  of  Three  Moments, 

4^+^  =  0; (i) 

m1  +  4m,  +  m3  =  0't (2) 

w*  +  4^3  +  ^4  =  o ;  .     ,     ,    ,     .     .     (3) 


mr.a  +  4mr.,  +  mr  =  —  w  -*  —  /)  =  —A,  suppose  ;    (r  —  i) 


r  +  mr+l  =  -w 


=  —  w  —(I  —  p)(2l  —/)=:—  B,  suppose  ;  (r) 
mr  +  4mr+1  +  mr+2  =  o  ;       .     .     .     .     (r  +  i) 


_i  =  O  ;      ....(«  —  2) 
=  o (n  —  i) 


4^0  THEORY   OF  STRUCTURES. 

By  equations  (i),  (2),  (3),  ...  (r  -  2), 

ill  I 


- 

the  upper  or  lower  sign  being  taken  according  as  r  is  even  or 
odd. 

By  equations  (n  —  i),  (n  —  2),  (n  —  3),  .  .  .  (r  +  i), 


.    mr+2  mr+l  mr 

=  -f  -  -  =  ±  - 


an-r-i 


The  coefficients  a  are  given  by  the  same  law  as  for  the  co- 
efficients a  in  Case  I.     Thus, 

ar  ...  an_r 

mr_2  =  —  --  mr_t     and     mr+l  =  —  -  mr. 


Substituting  these  values  of  mr_2  and  mr+l  in  the  (r  —  i)th 
and  rth  equations, 


and 


rtr-i  +  Mr\A  -  —^A  =  —  B  =  mr., 

an-r+2  ' 


where 

<*r-2 


and     c  =     ~ 


Hence,  solving  the  last  two  equations, 

Ac  -  B  Bb-A 

*r_I^-.— _    and     ^=_7-— . 

The  ratios  -    -  and   — ^-  are  each  less  than  unity,  and 

&r-i  &n-r+i 

hence  b  and  c  are  each  <  4  and  >  3. 


CONTINUOUS  GIRDERS.  481 

It  may  now  easily  be  shown  that  Ac  —  B  and  Bb  —  A  are 
each  positive.  Hence,  mr_^  and  mr  are  both  of  the  same  sign* 

The  bending  moment  mq  at  any  intermediate  support  on 
the  left  of  r  —  i  is  given  by 


mq  =  -|  --  —mr-i  if  9  and  r  are  the  one  even  and  the  other  odd, 


or 


=  --  --mr-i  if  9  an<3  r  are  both  even  or  both  odd. 


Thus  the  bending  moment  at  the  q\.\\  support  is  increased 
in  the  former  case  and  diminished  in  the  latter. 
If  q  is  on  the  right  of  r, 


mq  =  -\  —  *~q+l  mr  if  q  and  r  are  both  even  or  both  odd, 

an-r+i 

or 

mq-=.  --  ^^~mr  if  q  and  r  are  the  one  even  and  the  other  odd, 


and  the  bending  moment  on  the  ^th  support  is  increased  in  the 
former  case  and  diminished  in  the  latter. 

Thus,  the  general  principle  may  be  enunciated,  that,  "in  a 
horizontal  continuous  girder  of  n  equal  spans,  with  its  ends 
resting  upon  two  abutments,  the  bending  moment  at  an  inter- 
mediate support  is  greatest  when  the  two  spans  adjacent  to 
such  support,  and  the  alternate  spans  counting  in  both  direc- 
tions, are  loaded,  the  remainder  of  the  spans  being  unloaded." 

CASE  III.  The  same  general  principle  still  holds  true  when 
the  two  end  spans  are  of  different  lengths. 

E.g.,  let  the  length  of  the  first  span  be  kl,  k  being  a 
numerical  coefficient,  and  let  2(1  -\-  k)  =  x. 

Eq.  (i)  now  becomes 

m^x  -^m^  —  Q. 


482  THEORY  OF  STRUCTURES. 

Proceeding  as  before, 

.m^  _         m^  _     _m*  _     _  m,  _ 
}>'*      ~  V'      "  ^3  :      "  b,  ~~ 

the  coefficients  blt  b^,  b^,  .  .  .  being  given  by  the  same  law  as 
before,  viz., 


=..*•; 

,  =  4£3  -  ^  =  4*  —  i  ; 

4  =  4A  —  ^=  1  5*  -4J 

5  =  4^3  —  b%  =  56^—  15; 


The  two  sets  of  coefficients  (a)  and  (b)  are  identical  when 
x  =  4;  and  when  x  >  4,  all  the  coefficients  b  except  the  first 
(bl  —  i)  are  numerically  increased. 

Hence,  the  same  general  results  will  follow. 

N.B. — The  equations  giving  mq  are  simple  and  easily  ap- 
plicable in  practice.  They  may  be  written 

aq  B  —  Ac         . 

ma  =  ±  — -  -j—    —  if  q  is  on  the  left  of  r, 
ar_i    DC  —  I 

and 

mq  =  ±    "Ir  +  i  ~L —    -  if  ^  is  on  the  right  of  r. 

If  there  are  several  weights  on  the  rth  span, 
')    and    B  — 


EXAMPLE. — The  viaduct  over  the  Osse  consists  of  two  end 
spans,  each  of  94  ft.,  and  five  intermediate  spans,  each  of  126  ft. 
The  platform  is  carried  by  two  main  girders  which  are  con- 


MAXIMUM  BENDING  MOMENTS.  483 

tinuous  from  end  to  end.  The  total  dead  load  upon  the  girders 
may  be  taken  at  one  ton  (of  2000  Ibs.)  per  lineal  foot. 

Denote  the  supports,  taken  in  order,  by  the  letters  a,  b,  c,  d, 
e,f,g,  h,  and  let  it  be  required  to  find  the  maximum  bending 
moment  at  d  when  the  bridge  is  subjected  to  an  additional 
proof  load  of  ij  tons  per  lineal  foot. 

The  spans  ab,  cd,  de,  fgol  each  girder  carry  ij  tons  per 
lineal  foot. 

The  spans  be,  ef,  gh  of  each  girder  carry  \  ton  per  lineal  foot. 

Denoting  the  bending  moments  at  a,  b,  c,  d,  e,  ft  g,  h,  re- 
spectively, by  ml  ,;//„,...  ms,  the  intermediate  spans  by  /,  the 
end  spans  by  £/,  and  remembering  that  ml  =  o  =  m6  ,  we  have 

a  =  -~ 
4 

r 

,  +  4m3  +  m<  =  -  - 

r 

,  +  4mt  +  m,=  -  -(i  J-  +  i  J)  ; 


/a 

5  +  4^6  +  ^  =  -  -(4  + 


But  k  =  T9^-  =  f  ,  very  nearly. 


/a    2^ 

.-.  7m,  +  2m,  =  -  -  .  ^  ;    .....     (i) 

4     5 


-.-•      .....     (2) 


THEORY  OF  STRUCTURES. 


2mt  +  7m, 
From  eqs.  (i),  (2),  (3), 

From  eqs.  (4),  (5),  (6), 


ll  L. 

(V) 

4    2  ' 

/^       *7 

CA> 

4   2'      •     • 
£    7. 

CO 

4    2' 

(6} 

Hence,  ;«4,  the  maximum  required, 


16.  General  Theorem  of  Three  Moments. — The  most 
general  form  of  theorem  of  three  moments  may  be  deduced  as 
follows : 

Oi 02  03 


FIG.  334. 

Let  O,  X,  V,  the  (r  —  i)th,  rth,  and  (r  -\-  i)th  supports  of  a 
continuous  girder  of  several  spans,  be  depressed  the  vertical 
distances  d,  (=  O,O\  d,  (=O,X),  and  d,  (=  6>3F),  respectively, 
below  the  proper  level  Ofl^O^  of  the  girder. 


GENERAL    THEOREM   OF   THREE  MOMENTS.  48$ 

d^  ,  d^,  dz  are  necessarily  very  small  quantities. 

Let  OCXDVbe  the  deflection  curve,  and  let  the  tangent  at 
X  meet  the  verticals  through  O  and  V  in  E  and  F,  and  the 
tangents  at  O  and  V  in  Tt  and  7",  . 

Let  #,  be  the  change  of  curvature  from  O  to  X  (=  OTf). 
"  ^  "  "  "  "  "  "  Fto  X(=  FT^V). 

Let  Alf  A  i  be  the  effective  moment  areas  for  the  spans  OX, 
XV,  respectively. 

Let  .TJ  be  the  distance  (measured  horizontally)  of  the  centre 
of  gravity  of  Al  from  O. 

Let  x^  be  the  distance  (measured  horizontally)  of  the  centre 
of  gravity  of  A^  from  F. 

Let  0,E  =?>,  Of  =  y%. 

By  Art.  2, 


,         _ 
r         +I  h/,+J~^/1  4      "4+I/ 

But 

7i  -  <  _  <  -  J8          Z.I_L  A-^  .  ^» 

—7        —      7        >     °r      7*  "t"  7     ~~  /     IT    • 

/r  /r+I  /r  /r+I          /r  /r+x 

d^  —  d^d^  —  di        i  (A,x       A^\ 

-JT     /r+I  -^A  /r  -/—  ;• 

Again,  by  Art.  13,  Cor.  2, 

^ 

and 


^4r,  y4r+I  being  the  areas  of  the  bending-moment  curves  for 
the  spans  OX,  XV,  respectively,  on  the  assumption  that  they 
are  independent  girders,  or  cut  at  O,  X,  and  V,  and  zrj  zr^ 


486  THEORY  OF  STRUCTURES. 

being  the  horizontal  distances  of  the  centres  of  gravity  of  these 
areas  from  O  and  V.     Hence, 


Mr_Jr  +  2Mr(lr  +  /r+I)  +  Mr+llr+l 

KA*r       6  A      *r+l   -U 
.    =  —  bArj  --  vSlr+l-j  --  1- 

*r  tr+i 

Note.  —  If  O,  X,  or  V  is  above  O,O^  then  d,  ,  afa,  or  d^  is 
negative. 

Cor.  —  The  forms  of  the  Theorem  of  Three  Moments  given  in 
Cases  A  and  B,  Art.  13,  may  be,  immediately  deduced  from  the 
last  equation. 

CASE  A. 


A   --- 

^r- 


'   r+I 


CASE  B. 


17.  Advantages  and  Disadvantages  of  Continuous 
Girders.  —  The  advantages  claimed  for  continuous  girders  are 
facility  of  erection,  a  saving  in  the  flange  material,  and  the  re- 
moval of  a  portion  of  the  weight  from  the  centre  of  a  span  to- 


PROPERTIES   OF  CONTINUOUS   GIRDERS.  487 

wards  the  piers.  Circumstances,  however,  may  modify  these 
advantages,  and  even  render  them  completely  valueless.  The 
flange  stresses  are  governed  by  the  position  of  the  points  of  in- 
flexion, which,  under  a  moving  load,  will  fluctuate  through  a 
distance  dependent  upon  the  number  of  intermediate  supports 
and  upon  the  nature  of  the  loading.  In  bridges  in  which  the 
ratio  of  the  dead  load  to  the  live  load  is  small  the  fluctuation  is 
considerable,  so  that  for  a  sensible  length  of  the  main  girders, 
a  passing  train  will  subject  local  members  to  stresses  which  are 
alternately  positive  and  negative.  This  necessitates  a  local 
increase  of  material,  as.  each  member  must  be  designed  to  bear 
a  much  higher  stress  than  if  it  were  strained  in  one  way  only. 

Again,  the  web  of  a  continuous  girder,  even  under  a  uni- 
formly distributed  dead  load,  is  theoretically  heavier  than  if 
each  span  were  independent,  and  its  weight  is  still  further  in- 
creased when  it  has  to  resist  the  complex  stresses  induced  by 
a  moving  load. 

Hence,  in  such  bridges  the  slight  saving,  if  there  be  any, 
cannot  be  said  to  counterbalance  the  extra  labor  of  calculation 
and  workmanship. 

In  girders  subjected  to  a  dead  load  only,  and  in  bridges  in 
which  the  ratio  of  the  dead  load  to  the  live  load  is  large,  the 
saving  becomes  more  marked,  and  increases  with  the  number 
of  intermediate  supports,  being  theoretically  a  maximum  when 
the  number  is  infinite.  This  maximum  economy  may  be  ap- 
proximated to  in  practice  by  making  the  end  spans  about  four- 
fifths  the  intermediate  spans. 

In  the  calculations  relating  to  the  Theorem  of  Three  Mo- 
ments, it  has  been  assumed  that  the  quantity  El  is  constant, 
while  in  reality  E,  even  for  mild  steel,  may  vary  10  or  15  per 
cent  from  a  mean  value,  and  /  may  vary  still  more.  It  does 
not  appear,  however,  that  this  variation  has  any  appreciable 
effect  if  the  depth  of  the  girder  or  truss  changes  gradually,  but 
the  effect  may  become  very  marked  with  a  rapid  change  of 
depth,  as,  e.g.,  in  the  case  of  swing-bridges  of  the  triangular 


The  graphical  method  of  treatment  may  still  be  employed 
by  substituting,  for   the  actual   curve   of   moments,  a   reduced 


488  THEORY  OF  STRUCTURES. 

curve,  formed  by  changing  the  lengths  of  the  ordinates  in  the 
ratio  of  the  value  of  El  at  a  datum  section  to  EL 

It  is  often  found  economical  to  increase  the  depth  of  the 
girder  over  the  piers,  which  introduces  a  local  stiffness  and 
moves  the  points  of  inflexion  farther  from  the  supports.  A 
point  of  inflexion  may  be  made  to  travel  a  short  distance  by 
raising  or  depressing  one  of  the  supports. 

In  order  to  insure  the  full  advantage  of  continuity  the  ut- 
most care  and  skill  are  required  both  in  design  and  workman- 
ship.  Allowance  has  to  be  made  for  the  excessive  expansion 
and  contraction  due  to  changes  of  temperature,  and  the  piers 
and  abutments  must  be  of  the  strongest  and  best  description 
so  that  there  shall  be  no  settlement.  Indeed,  the  difficulties 
and  uncertainties  to  be  dealt  with  in  the  construction  of  con- 
tinuous girders  are  of  such  a  serious  if  not  insurmountable 
character  that  American  engineers  have  almost  entirely  dis- 
carded their  use  except  for  draw-spans. 

Much,  in  fact,  is  mere  guesswork,  and  it  is  usual  in  prac- 
tice to  be  guided  by  experience,  which  confines  the  points  of 
inflexion  within  certain  safe  limits. 

Under  these  circumstances  it  may  prove  desirable  to  fix 
the  points  of  inflexion  absolutely,  and  the  advantages  of  doing 
so  are  (a)  that  the  calculation  of  the  web  stresses  becomes 
easy  and  definite^  instead  of  being  complicated  and  even  in- 
determinate ;  (b)  that  reversed  stresses  (for  which  pin-trusses 
are  less  adapted  than  riveted  trusses)  are  almost  entirely 
avoided ;  (c)  that  the  stresses  are  not  sensibly  affected  by 
slight  inequalities  in  the  levels  of  the  supports ;  (d)  that  the 
straining  due  to  a  change  of  temperature  takes  place  under 
more  favorable  conditions. 

The  fixing  may  be  thus  effected  : 

(a)  A  hinge  may  be  introduced  at  the  selected  point. 

The  benefit  of  doing  so  is  very  obvious  when  circumstances 
require  a  wide  centre  span  and  two  short  side  spans. 

(b)  If  the  web  is  open,  i.e.,  lattice-work,  the  point  of  inflex- 
ion in  the  upper  flange  may  be  fixed  by  cutting  the  flange  at 
the  selected  point  and  lowering  one  of  the  supports  so  as  to 
produce  a   slight    opening   between  the   severed   parts.     The 


ADVANTAGES  OF  CONTINUOUS  GIRDERS.  489 

position  of  the  point  of  inflexion  in  the  lower  flange  is  then 
defined  by  the  condition  that  the  algebraic  sum  of  the  hori- 
zontal components  of  the  stresses  in  the  diagonals  intersected 
by  a  line  joining  the  two  points  of  inflexion  is  zero. 

It  must  be  remembered,  however,  that  this  fixing  of  the 
points  of  inflexion,  or  the  cutting  of  the  chords,  destroys  the 
property  of  continuity,  and,  indeed,  is  the  essential  distinction 
between  a  continuous  girder  and  a  cantilever. 

Four  methods  may  be  followed  in  the  erection  of  a  contin- 
uous girder,  viz.: 

1.  It  may  be  built  on  the  ground  and  lifted  into  place. 

2.  It  may  be  built  on  the  ground  and  rolled  endwise  over 
the  piers.     As  the  bridge  is  pushed  forward,  the  forward  end 
acts  as  a  cantilever  for  the  whole  length  of  a  span,  until  the 
next  pier  is  reached.     This  method  of  erection  is  common  in 
France. 

3.  It  may  be  built  in  position  on  a  scaffold. 

4.  Each  span  may  be  erected  separately,  and  continuity  pro- 
duced by  securely  jointing  consecutive  ends,  having  drawn  to- 
gether the  upper  flanges.     A  more  effective  distribution  of  the 
material  is  often  made  by  leaving  a  little   space  between   the 
flanges  and  forming  a  wedge-shaped  joint. 


49°  THEORY  OF  STRUCTURES. 


EXAMPLES. 

1.  Two  angle-irons,  each  2  in.  x  2  in.  x  £  in.,  were  placed  upon  sup- 
ports 12  ft.  9  in.  apart,  the  transverse  outside  distance  between  the  bars 
being  9^  in.,  and  were  prevented  from  turning  inwards  by  a  thin  plate 
upon  the  upper  faces.     The  bars  were  tested  under  uniformly  distributed 
loads,  and  each  was  found  to  have  deflected  2T5^  in.  when  the  load  over 
the  two  was  1008  Ibs.     Find  E  and  the  position  of  the  neutral  axis. 

Ans.  /=  J/£\  ;    E=  17,226,139   Ibs.;    neutral  axis  $£  in.  from* 
upper  face. 

2.  Both  bars  in  the  preceding  question  failed  together  when  the 
total  load  consisted  of  ioi  cwts.  (cwt.  =  112  Ibs.)  uniformly  distributed, 
and  3  cwts.  at  the  centre.     Find  the  maximum  stress  in  the  metal. 

Ans.  Compressive  unit  stress  =  20,323  Ibs. ; 
Tensile  unit  stress  =  39,577  Ibs. 

3.  Show  that  the  moments  of  resistance  of  an  elliptic  section  and  of 
the  strongest  rectangular  section  that  can  be  cut  out  of  the  same  are  in 
the  ratio  of  99  ^3  to  112,  and  that  the  areas  of  the  sections  are  in  the 
ratio  of  33  to  14  |/2. 

4.  Show  that  the  moments  of  resistance  of  an  isosceles  triangular 
section  and  of  the  strongest  rectangular  section  that  can  be  cut  out  of 
the  same  are  in  the  ratio  of  27  to  16,  and  that  the  areas  of  the  two 
sections  are  in  the  ratio  of  9  to  4. 

5.  An  angle-iron,   3  in.  x  3  in.  x  ^  in.,  was  placed   upon   supports 
12  ft.  9  in.  apart,  and  deflected  i-£  in.  under  a  load  of  8  cwts.  uniformly 
distributed  and  2  cwts.  at  the  centre.     Find  E  and  the  position  of  the 
neutral  axis. 

Ans.  E  =  16,079,611  Ibs.;  neutral  axis  iff  in.  from  upper  face. 

6.  The  effective  length  and  central  depth  of  a  cast-iron  girder  resting 
upon  two  supports  were  respectively  n  ft.  7  in.  and  10  in. ;  the  bottom 
flange  was  10  in.  wide  and  i  J  in.  thick ;  the  top  flange  was  2^  in.  wide 
and  •£  in.  thick;  the  thickness  of  the  web  was  f  in.     The  girder  was 
tested  by  being  loaded  at  points  3f  ft.  from  each  end,  and  failed  when 
the  load  at  each  point  was  17^  tons.      What  were   the   total   central 
flange  stresses  at  the  moment  of  rupture? 

What  was  the  central  deflection  when  the  load  at  each  point  was  7| 
tons?     (E  =  18,000,000  Ibs.,  and  the  weight  of  the  girder  =  3368  Ibs.) 

Ans.    164,747.4  Ibs.;  .368  in. 


EXAMPLES.  491 

7.  A  tubular  girder  rests  upon   supports  36  ft.  apart.     At  6  ft.  from 
one  end  the  flanges  are  each  27  in.  wide  and  2f  in.  thick,  the  net  area  of 
the  tension  flange  being  60  in.,  while  the  web  consists  of  two  fa-in. 
plates,  36  in.  deep  and  18  in.  apart.     Neglecting  the  effect  of  the  angle- 
irons  uniting  the  web  plates  to  the  flanges,  determine  the  moment  of 
resistance. 

The  girder  has  to  carry  a  uniformly  distributed  dead  load  of  56  tons, 
a  uniformly  distributed  live  load  of  54  tons,  and  a  local  load  at  the 
given  section  of  100  tons.  What  are  the  corresponding  flange  stresses 
per  square  inch? 

How  many  |--in.  rivets  are  required  at  the  given  section  to  unite  the 
angle-irons  to  the  flanges? 

Ans.  238.13  x  coeff.  of  strength  ;  3.3186  tons ;  3.896  tons. 

8.  A  yellow-pine  beam,  14  in.  wide  and  15  in.  deep,  was  placed  upon- 
supports  10  ft.  9  in.  apart,  and  deflected  f  in.  under  a  load  of  20  tons  at 
the  centre.     Find  E,  neglecting  the  weight  of  the  beam. 

Ans.  E  =  1,272,112  Ibs. 

9.  What  were  the  intensities  of  the  normal  and  tangential  stresses  at 
2  ft.  from  a  support  and  2\  in.  from  neutral  plane,  upon  a  plane  inclined 
at  30°  to  the  axis  of  the  beam  in  the  preceding  question? 

Ans.   132.83  and  218.91  Ibs. 

10.  A  beam  is  supported  at  the  ends  and  bends  under  its  own  weight. 
Show  that  the  upward  force  at  the  centre  which  will  exactly  neutralize 
the  bending  action  is  equal  to  f  or  \  of  the  weight  of  the  beam  (w)r 
according  as  the  ends  are  free  or  fixed. 

Find  the  neutralizing  forces  at  the  quarter  spans. 

Ans.  Ends  free  ^-gw  at  each  or  £faw  at  one  of  the  points  of 

division.  * 

Ends  fixed  ^w   at   each  or  %w  at  one  of   the  points  of 
division. 

11.  A  beam  8  in.  wide  and  weighing  50  Ibs.  per  cubic  foot  rests  upon 
supports  30  ft.  apart.      Find  its  depth  so  that  it  may  deflect  f  in.  under 
its  own  weight.     (E  —  i  ,200,000  Ibs.)  Ans.  9.185  in. 

12.  A  rectangular  girder  of  given  length   (/)  and  breadth  (&)  rests 
upon  two  supports  and  carries  a  weight  P  at  the  centre.     Find  its  depth 
so  that  the  elongation  of  the  lowest  fibres  may  be  l¥Vff  of  the  original 
length. 

Ans 

13.  A  yellow-pine  beam,  14  in.  wide,  15  in.  deep,  and  weighing  32  Ibs. 
per  cubic  foot,  was  placed  upon  supports   10  ft.  6  in.  apart.     Under 
uniformly  distributed  loads  of  59,734  Ibs.  and  of  127,606  Ibs.  the  central 


492  THEORY  OF  STRUCTURES. 

deflections  were  respectively  .18  in.  and  .29  in.     Find  the  mean  value 
of  E. 

Also  determine  the  additional  weight  at  the  centre  which  will  increase 
the  first  deflection  by  ^  of  an  inch.        Ans.  2,552,980  Ibs.;  24,121  Ibs. 

14.  In  the  preceding  question  find   for  the  load  of  59,734  Ibs.  the 
maximum  intensities  of  thrust,  tension,  and  shear  at  a  point   half-way 
between  the  neutral  axis  and  the  outside  skin  in  a  transverse  section  at 
one  of  the  points  of  trisection  of  the  beam.     Also  find  the  inclinations 
of  the  planes  of  principal  stress  at  the  point. 

Ans.  1609.255,  169.562,  119.364  Ibs. ;  0  =  3°  48!'. 

15.  A  pitch  pine  beam,  14  in.  wide,  15  in.  deep,  and  weighing  45  Ibs. 
per  cubic  foot,  is  placed  upon  supports  10  ft.  9  in.  apart,  and  carries  a 
load  of  20  tons  at  the  centre.     Find  the  deflection  and  curvature,  E 
being  1,270,000  Ibs.     What  stiffness  does  this  give  ? 

What  amount  of  uniformly  distributed  load  will  produce  the  same 
deflection?  Ans.  ^JT;  32  tons. 

16.  In  the  preceding  question  find  the  maximum  intensities  of  thrust, 
tension,  and  shear  at  points  (a)  half-way  between  the  neutral  axis  and 
the  outside  skin,  (b)  at  one  third  of  the  depth  of  the  beam,  in  a  trans- 
verse section  at  one  of  the  quarter  spans.     Also  find  the  inclinations  of 
the  planes  of  principal  stress  at  these  points. 

Ans.— (a)  95I-853>  292.969,  329.442  Ibs.;  6=9°  34$'. 
(b)  658.774,  171.108,  243.833  Ibs.;  6  =  15°  5of. 

17.  A  piece  of  greenheart,  142  in.  between  supports,  9  in.  deep,  and 
5  in.  wide,  was  tested  by  being  loaded  at  two  points,  distant  23  in.  from 
the  centre,  with  equal  weights.     Under  weights  at  each  point  of  4480 
Ibs.,  11,200  Ibs.,  and   17,920  Ibs.  the  central   deflections  were  .13  in.,  .37 
in.,  .67  in.,  respectively.     Find  the  mean  coefficient  of  elasticity.     The 
beam  broke    under    a   load   of   32,368   Ibs.    at  each   point.     Find   the 
coefficient  of  bending  strength. 

1 8.  A  sample  cast-iron  girder  for  the  Waterloo  Corn  Warehouses, 
Liverpool,  20  ft.  7-J-  in.  in  length  and  21  in.  in  depth  (total)  at  the  centre, 
was   placed   upon   supports    18   ft.    i£    in.   apart,   and   tested    under    a 
uniformly  distributed    load.     The    top   flange  was    5  in.  x  i£   in.,    the 
bottom  flange  was  18  in.  x  2  in.,  and  the  web  was   ij  in.  thick.     The 
girder  deflected  .15  in.,  .2  in.,  .25  in.,  and  .28  in.  under  loads  (including 
weight  of  girder)  of  63,763  Ibs.,  88,571  Ibs.,  107,468  Ibs.,  and  119,746  Ibs., 
respectively,  and  broke  during  a  sharp  frost  under  a  load  of  390,282  Ibs. 
Find  the  mean  coefficient  of  elasticity  and  the  central  flange  stresses  at 
the  moment  of  rupture. 

Ans.  7  =  3309.122;  E=  17,427,327  Ibs.;  20,121  Ibs.,  47,168  Ibs. 

19.  A  steel  rectangular  girder,  2  in.  wide,  4  in.  deep,  is  placed  upon 


EXAMPLES.  493 

supports  20  ft.  apart.     If  E  is  35,000,000  Ibs.,  find  the  weight  which,  if 
placed  at  the  centre,  will  cause  the  beam  to  deflect  i  in. 

Ans.   1296^7  Ibs. 

20.  A  timber  joist  weighing  48  Ibs  per  cubic  foot,  2  in.  wide  x  12  in. 
deep  x  14  ft.  long,  deflected  .825  in.  under  a  load  of  887  Ibs.  at  the 
centre.     Find  E.  Ans.  397,880  Ibs. 

21.  A  beam  of  span  /  is  uniformly  loaded.  Compare  its  strength  and 
stiffness  (a)  when  merely  resting  upon  supports  at  the  ends ;  (b)  when 
fixed  at  one  end  and  resting  upon  a  support  at  the  other ;  (c)  when  fixed 
at  both  ends.     In  case  (c)  two  hinges  are  introduced  at  points  distant  y 
from  the  centre ;  show  that  the  strength  of  the  beam  is  economized  to 

the  best  effect  when  y  = ,  and  that  the  stiffness  is  a  maximum  when 

y  =  —  very  nearly. 
4 


Ans.  Cases  (a)  and  (£).  mi  :m*m:i 
Cases  (a)  and  (c).  mi  :  ma  ::  3  :  2 
Case  (c).  Max.  economy,  mi  :  m* 


Max.   stiffness,  mi  :  m* 


:  D*  ::  i  :  .416. 
:  Da  ::  10  :  3. 


2  :  i 


5:24/2. 


4:3; 


15  :  4  (approx.). 

22.  A  beam  AB  of  span  /,  carrying  a  uniformly  distributed  load  of 
intensity  TV,  rests  upon  a  support  at  B  and  is  imperfectly  fixed  at  A,  so 

i   iif/a 
that  the  neutral  axis  at  A  has  a  slope  of  — g  -j=j .     The  end  B  is  lower 

than  A  by  an  amount  — jzj .     Find  the  reactions.     How  much  must  B 

be  lowered  so  that  the  whole  of  the  weight  may  be  borne  at  A  ?     Find 
the  work  done  in  bending  the  beam. 

21        ii  7  wl* 

Ans.   • — wl,  — ivl ;    -~  -==. 

32       32          48  £1 

23.  A  round  wrought-iron  bar  /  ft.  long  and  d  in.  in  diameter  can 
just  carry  its  own   weight.     Find  /  in  terms  of  d,  (a)  the  allowable  de- 
flection being  i  in.  per  100  ft.  of  span,  E  being  30,000,000  Ibs. ;  (ff)  the 
allowable  stress  being  8960  Ibs.  per  square  inch;  (V)  the  stiffness  given  by 
(a)  and  the  strength  given  by  (b)  being  of  equal  importance. 

Ans. — (d)  /=  |/2  50^? 2  ;    (b)  I  =  4/224^;    (c)  /=  \\d. 

24.  A  square  steel  bar  i  ft.  long  and  having  a  side  of  length  d  in.  can 
just  carry  its  own  weight;  its  stiffness  is  y^r  and  the  maximum  allow- 
able working  stress  is  7  tons  per  square  inch.     Find  /  in  terms  of  d,  E 
being  13,000  tons.  /(in  ft.)        13 

Ans.    -77: : — \   ==   — • 

<T(in  in.)         7 


494  THEORY  OF-  STRUCTURES. 

25.  A  uniformly  loaded   beam  with    both   ends   absolutely    fixed    is 
hinged  at  the  quarter-spans.     Show  that  the  slope  is  suddenly  doubled 
on  passing  a  hinge. 

26.  A  horizontal  beam  with  both  ends  absolutely  fixed  is  loaded  with 
a  weight  W  at  a  point  dividing  the  span  into  two  segments  a  and  b. 

W   (     ab     \3 
Show  that   the   deflection  at  the  point  is  —^.  I  j  ,  and  find  the 

work  done  in  bending  the  beam.  W*  I      ad 

HS' 


27.  Determine  the  isosceles  section  of  maximum  strength  which  can 
be  cut  out  of  a  circular  section  of  given   diameter,  and   compare   the 
strengths  of  the  two  sections. 

28.  A  3-in.  x  3-in.  x  £-in.  angle-iron,  with  both  ends  fixed  and  a  clear 
span  of  20  ft. .carries  a  uniformly  distributed  load  of  500  Ibs.  which  causes 
it  to  deflect  2  in.     Find  E.     What  single  load  at  the  centre  will  produce 
the  same  deflection  ?     Find  the  work  done  due  to  bending  in  each  case. 

Ans.  E  =  20,775,415  Ibs.;  250  Ibs. 

29.  A  steel  plate  beam  of  uniform  section  and  30  ft.  span  has  both 
ends  fixed  and  is  freely  hinged  at  the  points  of  trisection.     Determine 
the  neutral  axis(#)  for  a  uniformly  distributed  load  of  60,000  Ibs.;  (b)  for 
a  single  load  of  10,000  Ibs.  concentrated,^?™/1,  7^  ft.  and,  second,  15   ft. 
from  a  support. 

Ans.  (a)  Side   span,  y  =  -~=-  ( 100  —  Sx  H )  > 

/  N 

centre  span,^>  =  —  -f  -2 —  f  100  —  2o;r2  -f  jr8  j. 

(b)  First.   Loaded  span  between  support  and  weight, 


Loaded  span  between  weight  and  hinge, 
l25ar  ~  703125. 


Unloaded   side   span   horizontal  ;    centre   span 
straight  between  hinges. 


Second.  Side  span.jK  =  -  —  —  -f  %x  —  '—  )  ; 
El   \  6  / 


x2\       5000000 
centre  span,,  =  -5-  -j  + 


EXAMPLES.  495 

30.  A  uniformly  loaded  beam,  with  both  ends  absolutely  fixed,  is 
hinged  at  a  point  dividing  the  span  into  segments  a  and  b.  Draw  curves 
of  shearing  force  and  bending  moment,  and  compare  the  strength  and 
stiffness  of  the  beam  when  the  hinge  is  (a)  at  the  middle  point  ;  (b)  at 
a  point  of  trisection  ;  (c)  at  a  quarter-span.  Also,  determine  the  slope  of 
the  segments  of  these  points. 

w  $a*  +  8at>*  +  3^4  w  3#  4  4-  %a*b  +  5^4 

Ans'    *1=~  ~:    *2=-  ^ 


_  3^       */  _  wa  $a*  +  ^a*b  +4 

"  8    "       a3  +  P  '    8  0s  +  £8 

J/'  :  J/"  :  M'"  ::  14  :  14  :  ii  ;   £>'  :  £>''  :  £>'"::  6.25  :  3.29  :  2.66. 

Slopes  in  (#)  =  —  —  -  ;  in  (b)  =  —  -  —    for  segment  a, 
£  c  6  E  c  81 

and  =  —  —  —  -  for  segment  £  ; 

in  (c)  =  —  —  —f  for  segment  a,  and 
E  c  176 

/    /    92     r 

=  —  —  —  for  segment  b. 
EC  891 

31.  A  horizontal  beam  rests  upon  two  supports  and  is  loaded  with  a 
weight  W  at  a  point  dividing  the  span  into  segments  a  and  b.  Find 
the  deflection  at  this  point  and  the  work  done  in  bending  the  beam. 


W 

Ans- 


a*t>*          IV*    a*P  I      W  \ 

; =  —  x  deflection  . 

EI(a  +  b)     6EI  a  +  <H       2  / 


32.  A  wrought-iron  beam  of  rectangular  section  and   20  ft.  span  is 
16  in.  deep,  4  in.  wide,  and  is  loaded  with  a  proof  load  at  the  centre.     If 
the  proof  strength  is  7  tons  per  square  inch,  find  the  proof  deflection  and 
the  resilience,  E  being  12,000  tons.  Ans.  .029  ft.  ;  650  ft.-lbs. 

33.  Design  a  wooden  cantilever  12  ft.  long,  of  circular  section  and 
uniform  strength,  to  carry  a  uniformly  distributed  load  of  2  tons,  the 
coefficient  of  working  strength  being  i  ton  per  square  inch.     Also,  find 
the  deflection  of  the  free  end. 

Ans.  Taking  fixed  end  as  origin  and  z  being  radius  in  inches  at 
distance  x  ft.  from  origin,  then  \\z*  =  14(12  —  ;r)a. 

Deflection  at  end  =  — -^-  in. 
E 

34.  A  girder  fixed  at  both  ends  carries  (2n  -f-  i)  weights  ^concen- 
trated at  points  dividing  the  length  of  the  girder  into  2n  +  2  equal 

divisions.     Find  the  total  central  deflection.  n  +  i  Wl* 

Ans. =rr-. 

192     El 


THEORY  OF  STRUCTURES. 

35.  A  girder  30  m.  long  has  both  ends  fixed  and  carries  a  uniformly 
distributed  load  of  5800  k.  per  lineal  metre.     Find  the  deflection  and 

the  work  of  flexure.  567675000000 

Ans.  gj km. 

36.  A  steel  beam  of  circular  section  is  to  cross  a  span  of  15  ft.  and 
to  carry  a  load  of  10  tons  at  5  ft.  from  one  end.     Find  its  diameter,  the 
stiffness  being  such  that  the  ratio  of  maximum  deflection  to  span  is 
.00125.     E—  13,000  tons.  Ans.   10.3  in. 

37.  Determine  the   dimensions  of  a  beam   of   rectangular  section 
which  might  be  substituted  for  the  round  beam  in  the  preceding  ques- 
tion, the  stiffness  remaining  the  same  and  the  coefficient  of  working 
strength  being  7^  tons  per  square  inch.  Ans.  bd*  =  320. 

38.  The  flange  of  a  girder  consists  of  a  pair  of  angle-irons  and  of  a 
plate  which  extends  over  the  middle  portion  of  the  girder  for  a  certain 
required   distance.     Show  that  the  greatest   economy  of    material   is 
secured  when  the  length  of  the  plate  is  f  of  the  span  and  the  sectional 
areas  of  the   plate   and   angle-irons   are   as   4  to  5  (the  girder  being 
uniformly  loaded). 

39.  The   flange  of  a  uniformly  loaded   girder   is   to  consist  of  two- 
plates,  each  of  which  extends  over  the  middle  portion  of  the  girder  for 
a  certain  required  distance,  and  of  a  pair  of  angle-irons.     Show  that 
the  greatest  economy  of  material  is  realized  when  the  lengths  of  the 
plates  and  angle-irons  are  in  the  ratio  of  12  :  18  :  23,  and  when  the  areas 
of  the  plates  are  in  the  ratio  of  4  :  5. 

What  should  be  the  relative  lengths  of  the  plates  if  they  are  of  equal 
sectional  area  ?  AnSm   I  .  ^.  j(  ^  +  i). 

40.  An  elastic  beam  rests  upon  supports  at  its  ends,  and  a  weight 
placed   at   a   point  A  produces  a  certain  deflection  (d)  at  a   point  B. 
Show  that  if  the  weight  is  transferred  to  B  the  same  deflection  (d)  is 
produced  at  A. 

41.  A  uniform  beam  is  supported  by  four  equidistant  props,  of  which 
two  are  terminal.     Show  that  the  two  points  of  inflexion  in  the  middle 
segment  are  in  the  same  horizontal  plane  as  the  props. 

42.  Find  the  slope  and  deflection  at  the  free  end  of  the  following 
cantilevers  when  bending  under  their  own  weight,  /  being  the  length, 
2b  the  depth  at  the  fixed  end,  iv  the  specific  weight,  and  E  the  coefficient 
of  elasticity : 

(a)  Of  constant  thickness  /  and  with  profile  in  the  form  of  atrapezoid 
with  the  non-parallel  sides  equal  and  of  depth  20.  at  the  free  end. 

(b)  Of  circular  section  and  with  profile  in  the  form  of  an  isosceles 
triangle. 


EXAMPLES.  497 

(<r)  Of  constant  thickness  and  with  profile  in  the  form  of  a  parabola 
symmetrical  with  respect  to  the  axis  and  having  its  vertex  at  the  free 
end. 

wl*  yvl        (a*  a     P-  at>*  +  8a*6  -  2a*  )    ' 

;  rfv=*f  \  T^a  iog  'b  +  "    ~6F~    :  \  • 

wl*      i    wl*  2wl* 

' 


43.  Deduce  the  slope  and  deflection  at  the  free  end  — 

(d)  When  the  depth  2a  in  (a)  of  the  preceding  question  is  nil,  i.e., 
when  the  profile  is  an  isosceles  triangle. 

(e)  Due   to   a   uniformly   distributed   load   of  intensity  p  over  the 
cantilever  (a).     Hence,  also,  deduce  the  slope  and  deflection  when  the 
depth  2a  is  nil. 

(/)  Due  to  a  weight  W  at  the  free  end  of  (a). 

(g)  Due  to  a  uniformly  distributed  load   of   intensity  p  upon   the 
cantilever  (<:). 

Wl*       Wl* 


a      (2?  +  $ab  -  a*)(b  -  a) 


) 

I90' 


3  pl* 
~4~EtP' 


^'°g«      3*-« 


b-a 


2  EPt*    10 


44.  A  cantilever  of  fength  /,  specific  weight  -w,  and  square  in  section, 
a  side  of  the  section  being  2b  at  the  fixed  and  2a  at  the  free  end,  bends 
under  its  own  weight.     Find  the  slope  and  deflection  of  the  neutral  axis 
at  the  free  end.     Hence,  also,  deduce  corresponding  results  when  the 
cantilever  is  a  regular  pyramid. 

(b 

45.  If  the  section  of  the  cantilever  in  the  preceding  question,  instead 
of  being  square,  is  a  regular  figure  with  any  number  of  equal  sides,  show 
that  the  neutral  axis  is  a  parabola  with  its  vertex  at  the  point  of  fixture. 

46.  The  section  of  a  cantilever  of  length  /  is  an  ellipse,  the  major  axis 
(vertical}  being  twice  the  minor  axis.     Find  the  deflection  at  the  end 


49^  THEORY   OF  STRUCTURES. 

under  a  single  weight  W,f  being  the  coefficient  of  working  strength 
and  E  the  coefficient  of  elasticity.  /  297     /V5   \  J 

Ans.  ---  £m?7T 

\7ooo  h  3  W  I 

47.  A  cast-iron  beam  of  an  inverted  T-section  rests  upon  supports 
22  ft.  apart;  the  web  is  I  in.  thick  and  20  in.  deep;  the  flange  is  1.2  in. 
thick  and  12  in.  wide;  the  beam  carries  a  uniformly  distributed  load  of 
99,000  Ibs.     Find  the  maximum  deflection,  E  being  17,920,000  Ibs. 

Ans.  .822.  in.  (/  =  1608.65). 

48.  Find  the  maximum  deflection  of  a  cast-iron  cantilever  2  in.  wide 
x  3  in.  deep  x  120  in.  long  under  its  own  weight,  E  being  17,920,000  Ibs. 

Ans.   H  in. 

49.  A  girder  of  uniform,  strength,  of  length  /,  breadth  b,  and  depth  d, 
rests  upon  two  supports  and  carries  a  uniformly  distributed  load  of  w  Ibs. 
per  unit  of  length,  which  produces  an  inch-stress  of  /  Ibs.  at  every  point 

n  —  2   f\t  b  \\ 

of  the  material.     Show  that  the  central  deflection  is  --  ™(~       /t 

2      E  \  3«y 

when  b  is  constant  and  ^variable.     Find  the  deflection  when  d  is  con- 

stant and  b  variable.  fr 

Ans.  +—. 
4,Ed 

50.  A  semi-girder  of  uniform  strength,  of  length  /,  breadth  b,  and 
depth  d,  carries  a  weight  W  at  the  free  end  which  produces  an  inch- 
stress  of/  Ibs.  at  every  point  of  the  material.     Prove  that  the  maximum 

deflection  is  —  -  —  [  —  -)  when  b  is  constant  and  d  variable,  and  that 
3     E   \6  W) 

it  is  twice  as  great  as  it  would  be  if  the  section  were  uniform  throughout 
and  equal  to  that  at  the  support. 

What  would   be  the    maximum    deflection    if  the  semi-girder  were 
subjected  to  a  uniformly  distributed  load  of  w  Ibs.  per  unit  of  length  ? 


Ans.  . 

EJ     $w 

51.  The  neutral  axis  of  a  symmetrically  loaded  girder,  whose  moment 
of  inertia  is  constant,  assumes  the  form  of  an  elliptic  or  circular  arc. 
Show  that  the  bending  moment  at  any  point  of  the  deflected  girder  is 
inversely  proportional  to  the  cube  of  the  vertical  distance  between  the 
point  and  the  centre  of  the  ellipse  or  circle. 

52.  A   vertical   row   of    water-tight    sheet    piling,     12    ft.    high,    is 
supported   by   a   series   of   uprights   placed  6  ft.  centre  to  centre  and 
securely  fixed  at  the  base.     Find  the  greatest  deviation  of  an  upright 
from  the  vertical  when  the  water  rises  to  the  top  of  the  piling.     What 
will  the  maximum  deviation  be  when  the  water  is  6  ft.  from  the  top  ? 

wblf        3110400        wb  wbc  218720 

Ans.  —-  =  —~—  :    —=r.  (h  -  cY  +  —=-,(*  -  cY  =  -~-  . 
3o£7  El         y>EI  24  El  hi 


EXAMPLES.  499 

53.  A  vertical  row  of  water-tight  sheet  piling,  30  ft.  high,  is  supported 
by  a  series  of  uprights  placed  8  ft.  centre  to  centre  and  securely  fixed  at 
the  base,  while  the  upper  ends  are  kept  in  the  vertical  by  struts  sloping 
at  45°.  If  the  water  rises  to  the  top  of  the  piling,  find  (a)  the  thrust  on  a 
strut  ;  (d)  the  maximum  intensity  of  stress  in  an  upright  ;  (c)  the  amount 
and  position  of  the  maximum  deviation  of  an  upright  from  the  vertical. 

iuh* 
Ans.  45000  |/2  Ibs.  ;  max.  B.  M.  =  --  —  ,  and  max.  intensity  of 


stress  =  —  .  1  -  -  ±  -7-  --  —   -  ;   deflection   is  a  max.   when 

A  (    10        /  I5|/5   j 

h  30  7vh*      32 

x  =  —  —  =  —  •=,  and  its  amount  =  —^j  ----  -  . 

V5          VS  ^  7S°V5 

54.  The  piling  in  the  preceding  example  is  strengthened  by  a  second 
series  of  struts  sloping  at  45°  from  the  points  of  maximum  deviation. 
Find  the  normal  reactions  upon  an  upright  and  the  bending  moment  at 
its  foot. 

What  will  be  the  reactions  and  bending  moment  if  the  second  row  of 
struts  starts  from  the  middle  of  the  uprights? 

Ans.  .00754W/*2  ;  .lyjwk*  ;  .^2O27iw/t3;     -^wk*  ;  If^rf*  ;  fH^//9. 

55.  A  continuous  girder  of  three  spans,   the   outside   spans   being 
equal,  is  uniformly  loaded.     What  must  be  the  ratio  of  the  lengths  of 
the  centre  and  a  side  span  so  that  the  neutral  axis  may  be  horizontal 
over  the  intermediate  supports?  Ans.    4/T  :  y'T. 

56.  What  should  the  ratio  be  if  the  centre  span  is  hinged  (a)  at  the 
centre;  (ff)  at  the  points  of  trisection  ?  Ans.—  (a)  4/2"  :  i  ;  (£)  3  :  2  \/~2. 

57.  Four  weights,  each  of  6  tons,  follow  each  other  at  fixed  distances 
of  5  ft.  over  a  continuous  girder  of  two  spans,  each  equal  to  50  ft.    If  the 
second  and  third  supports  are   i    in.  and   i£  in.,  respectively,  vertically 
below  the  first  support,  find  the  maximum  B.  M.  at  the  intermediate 

support.  /  El   \  , 

Ans.     .9855  --  1  ft.  -tons. 
\  40000,1 

58.  A  continuous  girder  of  two  equal  5o-ft.  spans  is  fixed  at  one  of 
the  end  supports.     The  girder  carries  a  uniformly  distributed  load  of 
loco  Ibs.  per  lineal  foot.     Find  the  reactions  and  bending  moments  at 
the  points  of  support.     How  much  must  the   intermediate  support  be 
lowered  so  that  it  may  bear  none  of  the  load  ?     How  much  should  the 
free  end  be  then  lowered  to  bring  upon  the  supports  the  same  loads  as  at 
the  first  ? 

Ans.  Reactions  =  23,214^,  57,142^,  19,642!  Ibs.  ; 

Bending  moments  =  178,571^,  267,8574  ft.  -Ibs; 


500  THEORY  OF  STRUCTURES. 

59.  Four  loads,  each  of  12  tons  and  spaced  5,4,  and  5  ft.  apart,  travel 
in  order  over  a  continuous  girder  of  two  spans,  the  one  of  30  and  the  other 
of  20  ft.     Place  the  wheels  so  as  to  throw  a  maximum  B.  M.  upon  the 
centre  support,  and  find  the  corresponding  reactions. 

Draw  a  diagram  of  B.  M.,  and  find  the  maximum  deflection  of  each 
span. 

60.  The  loads  upon  the  wheels  of  a  truck,  locomotive,  and  tender, 
counting  in  order  from  the  front,  are  7,  7,  10,  10,  10,  10,  8,  8,  8,  8  tons, 
the  intervals  being  5,  5,   5,   5,  5,  9,  5,  4,  5  ft.     The  loads  travel  over  a 
continuous  girder  of  two  5o-ft.  spans  AB,  BC.     Place  the  locomotive, 
etc.,  (a)  on  the  span  AB  so  as  to  give  a  maximum  B.  M.  at  B\  (8)  so  as 
to  give  an  absolute  maximum  B.  M.  at  B. 

61.  A  continuous  girder  of  two  spans  AB,  BC  has  its  two  ends  A 
and  C fixed  to  the  abutments.     The  load  upon  AB  is  a  weight  P  distant 
/  from  A,  and  that  upon  BC  a  weight  Q  distant  g  from  C.     The  length 
of  AB  =  l\ ,  of  BC  —  /a .     The  bending  moments  at  A,  B,  C  are  Mi ,  M* , 
Ms,  respectively.     The  areas  of  the  bending-moment  curves  for  the 
spans  AB,  BC  assumed  to  be  independent  girders  are^i,  At,  respect- 
ively.    Show  that 

Mill  +  Mt(li  +  /a)  +  M*h  =  -  2(Ai  +  A*), 
and  M*(li  +  /,)  =  — 

If  /,  =  /3  =  /,  show  that  Mi  is  a  maximum  if 
2l(Pp  -  Qg)  = 

62.  A  continuous  girder  of  two  spans  AB,  BC  rests  upon  supports  at 
A,  B.     A  uniformly  distributed  load  EF  travels  over  the  girder.     G\  is 
the  centre  of  gravity  of  the  portion  BE  upon  AB,  and  G*  that  of  the 
portion./?/''  upon  BC.     If  the  bending  moment  at  B  is  a  maximum,  show 
that 

AE.EB  _  Ad 
CF  .  FB  ~  CG*' 

63.  An  eight-wheel  locomotive  travels  over  a  continuous  girder  of 
two  loo-ft.  spans ;  the  truck-wheels  are  6  ft.  centre  to  centre,  the  load 
upon  each  pair  being  8000  Ibs. ;  the  driving-wheels  are  8£  ft.  centre  to 
centre,  the  load  upon  each  pair  being  16,000  Ibs.  ;  the  distance  centre  to 
centre  between  the  front  drivers  and  the  nearest  truck-wheels  is  also 
8£  ft.     Place  the  locomotive  so  as  to  throw  a  maximum  B.  M.  upon  the 
centre  support,  and  find  the  corresponding  reactions. 

64.  If  an  end  of  a  continuous  girder  of  any  number  of  spans  is  fixed, 
show  that  the  relation  between  the  moment  of  fixture  (Mi)  and  the 

W/2 

bending  moment  (M*)  at  the  consecutive  support,  is  2Mi  +  M?  =  — — , 

4 

or  2J/i  +  M*  =  —  ~^S[Pp(l  —  p)(2l  —  /)],  according  as  the  load    upon 


EXAMPLES.  5OI 

the  span  (/)  between  the  fixed  end  and  the  consecutive  support  is  of 
uniform  intensity  or  consists  of  a  number  of  weights  Pi,  P*,  Pa,  .  .  . 
concentrated  at  points  distant  A,  /«,  /s,  .  .  .  from  the  fixed  end. 

65.  A  continuous  girder  of  two  spans  AB,  BC,  carrying  a  load  of 
uniform  intensity,  has  one  end  A  fixed,  and  the  other  end  rests  upon  the 
support  at  C.     If  the  bending  moments  at  A  and  B  are  equal,  show  that 
the  spans  are  in  the  ratio  of  1/3  to  f/2,  and  find  the  reactions  at  the 
supports,  Wi  being  the  load  upon  AB,  and  W*  that  upon  BC. 

Ans.  At  A  reaction  = 
«  B  "  = 
««  C  "  =tJF.. 

66.  A  viaduct  over  the  Garonne  at  Bordeaux  consists  of  seven  spans, 
viz.,  two  end  spans,  each  of  57.375  m.,  and  five  intermediate  spans,  each 
of  77.06  m.  ;  the  mam  girders  are  continuous  from  end  to  end,  and  are 
each  subjected  to  a  dead  load  of  3050  k.  per  lineal  metre.     Determine  the 
absolute  maximum  bending  moment  at  the  third  support  from  one  end. 
Also  find  the  corresponding  reactions,  the  points  of  inflexion,  and  the 
maximum  deflection  in  the  first  and  second  spans. 

67.  A  continuous  girder  consists  of  two  spans,  each  50  ft.  in  length; 
the  effective  depth  of  the  girder  is  8  ft.     If  one  of  the  end  bearings 
settles  to  the  extent  of  i  in.,  find  the  maximum  increase  in  the  flange 
and  shearing  stress  caused  thereby,  and  show  by  a  diagram  the  change 
in  the  distribution  of  the  stresses  throughout  the  girder.     (Assume  the 
section  of  the  girder  to  be  uniform,  and  take  E  =  25,000,000  Ibs.) 

Ans.  Increase  of  maximum  B.  M.  =  s-A£f[ — ~f i  I, 

\2i6rf        / 

"  shearing  force       =  ff/, 
w  being  weight  per  unit  of  length,  and  /the  moment  of  inertia. 

68.  A  girder  carrying  a  uniformly  distributed  load  is  continuous  over 
four  supports,  and  consists  of  a  centre  span   (/2)  and   two  equal  side 
spans  (/i).       Find  the  ratio  of  l\  to  /2 ,  so  that  the  neutral  axis  at  the 
intermediate  supports  may  be  horizontal.     Also  find  the  value  of  the 
ratio  when  a  hinge  is  introduced  (a)  at  the  middle  point  of  the  centre 
span  ;  (fr)  at  the  points  of  trisection  of  the  centre  span  ;  (c)  at  the  middle 
points  of  the  half  lengths  of  the  centre  span. 

A       £I_i    ^-!    £!_!    A'-3 
'    /,«-3'   /22~2'/2'-9;^~4' 

69.  In  a  certain  Howe  truss  bridge  of  eight  panels,  the  timber  cross- 
ties  are  directly  supported    by  the   lower   chords,  and   are  placed   suffi- 
ciently close  to  distribute  the  load  in  an  approximately  uniform  manner 
over  the  whole  length  of  these  chords,  thus   producing   an    additional 
stress  due  to  flexure.     Assuming  that  the  chords  may  be  regarded  as 
girders  supported  at  the  ends  and   continuous   over  seven   intermediate 


5O2 


THEORY  OF  STRUCTURES. 


supports  coincident  with  the  panel  points,  and  that  these  panel  points 
are  in  a  truly  horizontal  line,  determine  (a)  the  bending  moments  and 
reactions  at  the  panel  points  ;  (b)  the  maximum  intermediate  bending 
moments;  and  (c)  the  points  of  inflection,  corresponding  to  a  load  of  w 
per  unit  of  length,  /  being  the  length  of  a  panel. 

Ans.  —  (a)  At  i  st  support  ;  2d   support;  3d    support; 

B.  M.  =  o  ;      -¥VW2; 

Reaction  =  R,  =  £     a//  ;  R*  =        ivl  ;  Rz  = 


At  4th  support  ;  5th  support. 

B.  M.  =  -^W2:  -H^l\ 

Reaction  =  Ri=  fffw/;     Rs>  —  f  ffw/. 


Maximum  intermediate  B.  M.  =  -  '- 


in  istspan  ; 


6208.  5 

=  W"/     4 

(<:)  Points  of   inflexion   in  the  four  spans   are   given    by 
*  =  —  =  ^/;  *i(/  +  *) 

W  388 


x) 
x) 


-*02  =  o  ; 
xf  =  o. 


70.  A.  continuous  girder  of  two  equal  spans  \&fixefl  $&  one  of  the  end 
supports.  The  girder  carries  a  uniformly  distributed  load  of  intensity 
w.  If  the  length  of  each  span  is  /.  find  the  reactions  and  moment  of 
fixture.  How  much  must  the  intermediate  support  be  lowered  so  that 
it  may  bear  none  of  the  load  ?  How  much  should  the  free-end  support 
then  be  lowered  to  bring  upon  the  supports  the  same  loads  as  before  ? 

ii        16       13  w/2      c  wl*      5  «//* 

—       —  wl  —tvl   —  —  •    -  —  —     —  —— 


Ans.  —wl,  —  wl,  —tvl\ 
28       14       28 


14 


24 


71.  Each  of  the  main  girders  of  the  Torksey  Bridge  is  continuous 
and  consists  of  two  equal  spans,  each  130  ft.  long.  The  girders  are 
double-webbed  ;  the  thickness  of  each  web  plate  is  i  in.  at  the  centre 
and  f  in.  at  the  abutments  and  centre  pier  ;  the  total  depth  of  the  gir- 
ders is  10  ft.,  and  the  depth  from  centre  to  centre  of  the  flanges  is  9  ft. 
4§  in.---FTtidX/?)  the  reactions  at  the  supports,  and  also  (b)  the  points  of 
inflexion,  when  2oc\tpns  of  live  load  cover  one  span,  the  total  dead  load 
upon  each  span  being.  180  tons  uniformly  distributed.  The  top  flange  is 


EXAMPLES. 


503 


cellular  ;  its  gross  sectional  area  at  the  centre  of  each  span  is  51  sq  in  , 
and  the  corresponding  net  sectional  area  of  the  bottom  flange  is  55  sq. 
in.  Determine  (6-)  the  flange  stresses  and  (d}  the  position  of  the  neutral 
axis.  (7=372,500.)  Also  (e)  determine  the  reactions  when, first,  B  and, 
second,  Care  lowered  i  in.  (E '=  16,900  tons.) 
Ans. — (a)  155,  350,  55  tons. 

(£)  io6r^  and  79!  ft.  from  end  support. 

(c)  6  7    and    7.3   tons  per  sq.  in.  in   loaded  span;    1.13   and 

1.22  tons  per  sq.  in.  in  unloaded  span. 

(d)  58.3  in.  from  centre  line  of  top  flange. 


(e)  First.  R,  = 


i  EI 


Second.  Ri  =  1 55  —  ~  ^TT 


lEI 

=  350-  sTTi 
I    £7 

-S?+;T- 

lEI 

=  350+--^-; 
4  * 

i  EI 
'=55-o  •»• 


Where 


}.£/ 


18625 
11232 


72.  Two  tracks,  6  ft.  apart,  cross  the  Torksey  Bridge,  and  are  sup- 
ported by  single-webbed  plate  cross-girders  25  ft.  long  and  14  in.  deep. 
If  the  whole  of  the  weight  upon  a  pair  of  drivers,  viz.,  10  tons,  be  directly 
transmitted  to  one  of  these  cross-girders,  draw  the  corresponding  shear- 
ing-force and  bending-moment  diagrams  (i)  if  the  ends  of  the  cross- 
girder are  fixed  to  the  bottom  flanges  of  the  main  girders  ;  (2)  if  they 
merely  rest  on  the  said  flanges.  Find  the  maximum  deflection  of  the 
cross-girder  and  the  work  done  in  bending  it,  in  each  case. 


Ans.  (i) 


-  at  13.208  ft.  from  one  end. 
EI 

Total  work  of  flexure  =          .--  ft  -tons. 


73.  A  swing-bridge  consists  of  the  tail  end  AB,  and  of  a  span  ffC,  of 
length  i  ft.,  the  pivot  being  at  B.  The  ballast-box  of  weight  ?Fextends 
over  a  length  AD  (=  2c  ft.),  and  the  weight  of  the  bridge  from  D  to  B 
is  w  tons  per  lineal  foot.  If  DB  =  x,  if  p  is  the  cost  per  ton  of  the 
bridge,  and  if  q  is  the  cost  per  ton  of  the  ballast,  show  that  the  total  cost 

is  a  minimum   when   x  +  c  =  (- — ),  and  that  the  corresponding 

weight  of  the  ballast  is  wx(  —   -  i  J 


504  THEORY  OF  STRUCTURES. 

74  Compare,  graphically,  the  shearing  forces  and  bending  moments 
along  the  span  BC  of  the  bridge  in  the  preceding  question  when  the 
bridge  is  closed,  with  their  values  when  the  bridge  is  open.  What  pro- 
vision should  be  made  to  meet  the  change  in  the  kind  of  stress  ? 

75.  Each  of  the  main  girders  of  a  railway  bridge  resting  upon  two 
end  supports  and  five  intermediate  supports  is  fixed  at  the  centre  sup- 
port, is  3  ft.  deep  throughout,  and  is  designed  to  carry  a  uniformly  dis- 
tributed dead  load  of  t  ton  and  a  live  load  of  |  ton  per  lineal  foot.  The 
end  spans  are  each  51  ft.  8  in.  and  the  intermediate  spans  each  50  ft.  in 
the  clear.  Find  the  reactions  at  the  supports.  The  girders  are  single- 
webbed  and  double-flanged  ;  the  flanges  are  12  in.  wide  and  equal  in 
sectional  area,  the  areas  for  the  intermediate  spans  being  13  sq.  in.  and 
17  sq.  in.  at  the  centre  and  piers  respectively.  Find  the  corresponding 
moments  of  resistance  and  flange  stresses,  the  web  being  £  in.  thick. 

Ans.  Reaction  at  istand  7th  supports  =  I5HMM  »  at  2C^  an^  5th 
supports  =  43fff  til;  at  3d  and  5th  supports  =  35|ffl  ; 
at  4th  support  =  3%'?/4\8s  tons. 

At  piers—  =  693  and  flange  stresses  are  3.59  tons  per  sq.  in. 

at  2d  support,  2.45  at  3d,  and  2.83  at  4th. 
At  centre  —==  549  and  flange  stresses  in  istspan  =  3.2  tons 

per  sq.  in.,  in  2d  =  1.3,  and  in  3d  =  1.78. 

76.  A  continuous  beam  of  four  equal  spans  carries  a  uniformly  dis- 
tributed load  of  w  intensity  per  unit  of  length.  The  second  support  is 
depressed  a  certain  distance  d  below  the  horizontal,  and  the  reaction  at 
the  2d  support  is  twice  that  at  the  ist.  Show  that  the  reactions  at  the 
ist,  2d,  3d,  4th,  and  5th  supports  are  in  the  ratio  of  the  numbers  15,  30, 
36,  34,  and  1  3  ;  find  d.  With  this  same  value  of  d  find  the  reactions  when 
one  end  infixed. 

i  wP 


Ri  = 

77.  A  continuous  girder  of  two  equal  spans  (/)  is  uniformly  loaded. 
Show  that  the  ends  will  just  touch  their  supports  if  the  centre  support 

w/4 
is  raised  —  • 

78.  If  ^i  ,  di  ,  d3  ,  di  are  respectively  the  deflections  of  the  ist,  2d,  3d, 
and  4th  panel  points  in  question  69,  show  that  the  bending  moment  at 
the  middle  panel  point  (J/4)  is  given  by 


EXAMPLES.  505 

79.  A  girder  supported  at  the  ends  is  30  ft.  in  the  clear  and   carries 
two  stationary  loads,  viz.,  7  tons  concentrated  at  6  ft.  and  12  tons  at  18 
ft.  from  the  left  support.     Find  the  position  and  amount  of  the  maxi- 
mum deflection,  and  also  the  work  of  flexure.     The  girder  is  built  up  of 
plates  and  angle-irons  and  is  24  in.  deep.     If  the  moment  of  resistance 
due  to  the  web  is  neglected,  and  if  the  intensity  of  the  longitudinal  stress 
is  not  to  exceed  5  tons  per  sq.  in., _  what  should  be  the  flange  sectional 
area  corresponding  to  the  maximum  bending  moment. 

Ans.  Max.  deflection  =  f fr8  —  %(x  —  6)3  —  ^g&x,  where 
x  =  1 5.34  ft. 

67161.6  , 
Work  =  ----;,_— ft.-tons. 

J^LJ. 

Sect,  area  =  10.32  sq.  in. 

80.  Determine  the  work  of  flexure  and  the  necessary  flange  sectional 
area  at  the  centre  if  the  girder  in  the  preceding   question  is  subjected 
to  a  uniformly  distributed  load  of  40  tons,  instead  of  the  isolated  loads. 

1540000, 
Ans.  Work  =  - — - — ft. -tons;  sect,  area  =  15  sq.  in, 

81.  (a)  The  bridge  over  the  Garonne  at  Langon  carries  a  double 
track,  is  about  695  ft.  in  length,  and  consists  of  three  spans,  AB,  BC,  CD. 
The  two  main  girders  are  continuous  and  rest  upon  the  abutments  at 
A  and  D  and  upon  piers  at  B  and  C.     The  effective  length  of  each  of 
the  spans  AB,  CD  is  208  ft.  6  in.,  and  of  the  centre  span  J3C  24.3  ft.     The 
permanent  load  upon  a  main  girder  is  1277  Ibs.  per  lineal  foot,  and  the 
proof  load  is  2688  Ibs.  per  lineal  foot.     Find  the  reactions  at  the  sup- 
ports (i)  when  the  proof  load  covers  the  span  AB ;  (2)  when  the  proof 
load  covers  the  span  BC ';  (3)  when  the  proof  load  cover  the  spans  AB 
and  BC ';  (4)  when  the  proof  load  covers  the  whole  girder. 

Draw  shearing-force  and  bending-moment  diagrams  for  each  case. 

(b)  At  the  piers  the  web  is  £  in.  thick  and  18  ft.  in  depth,  and  each 
flange  is  made  up  of  four  plates  £  in.  thick  and  3  ft.  wide.     Determine 
the  flange  stresses  for  cases  (i)  and  (3). 

(c)  The  angle-irons  connecting  the  flanges  with  the  web  at  the  pier 
are  riveted  to  the  former  with  i^-in.  rivets  and  to  the  latter  with  i-in. 
rivets.     How  many  of  each  kind  are  required  in  one  line  per  lineal  foot 
on  both  sides  of  the  pier  at  B,  8000  Ibs.  per  square  inch  being   safe 
shearing  stress  ? 

(d)  The  effective  height  of  the  pier  at  B  is  41  ft.,  its  mean  thickness 
is  14  ft.  9  in.,  its  width  is  42  ft.  9  in.,  and  it  weighs  125  Ibs.  per  cubic 
foot.     If  there  is  no  surcharge  on  the  bridge,  and  if  the  coefficient  of 
friction  between  the  sliding  surfaces  at  the  top  of  the  pier  is  taken  at 
.15,  show  that  the  overturning  moment   due  to  the  dilatation  of  the 
girders  is  about  -fa  of  the  amount  of  stability  of  the  pier. 


506  THEORY  OF  STRUCTURES. 

(e)  Find  the  points  of  inflexion  and  also  the  maximum  deflections  in 
Case  3. 

What  practical  advantage  is  derived  from    the  calculation    of  the 
deflection  ? 

Ans.—(a)    Case  i.  R\  =  353469.95;  R?  =  656955.7; 

Rs  =  280612.55  ;  Ri  =  109608.77  Ibs. ; 
Mi=  — 12247115.3;  M3  =  —  4823424.5  ft.-lbs. 
Case  2.  R,  =  68185  2  =  R,  Ibs.  ; 
A>2  =  6791783  =  ^?3  Ibs.; 
M*=  —  13439537.7  ft.-lbs.  =  Aft. 
Case  3.  RI  =  312982.65  ;  7?2  =  1024035  ; 

Rs  =  647691  ;        Rt  =  69121.47  Ibs.  ; 
Mi=  —  1565031.2 ;  Aft  =  —  8226621.2  ft.-lbs. 
Case  4.  .#1=422591.42    =y?4lbs.; 
7?2  =  1304647. 55  =  R3  Ibs. ; 
J/2=  —  15455566  =  J/a  ft.-lbs. 

(0)   /=  2130816;  in  case  i,/2  =  7448.9  Ibs. per sq.  in. 

/»  =  2933-6    "      "    "    " 
in  case  3,/2  =  9400.3    "      "    "    " 

73  =  5003.5    "      "    "    " 
(Weakening  effect  of  rivet-holes  in  tension  flange  is 

neglected.) 
(?)    9.1  per  lineal  foot;  11.5  pe^r  lineal  foot. 

(d)  Moment  of  stability   =  23833291^11  ft.-lbs. ; 
overturning  moment  =  1919408.8  ft.-lbs.; 

ratio  =  12.4. 

(e)  Points  of  inflexion:   in  AB,  157.8  ft.  from  A',   in  BC, 

at  a  distance  x  from  B  given  by  x*— 2581^+10426^ 
=  o;  in  CD,  at  54.1  ft.  from  D. 
Max.  deflections: 

In  AB,        —  (165/5 jr4—  52l63-7*8  +  227693091.6*), 

where*  is  given  by  66o|*2 — 156491. 3*  + 227693091. 6=0  ; 
In  BC,  —7(1 65.2**— 853829*"  +  9977485. 9*2— 10327286968), 

xS/ 

where  *  is  given  by  *2  —  3876*  +  30196^  =  o. 

82.  A  beam  AB  of  span  /  carrying  a  uniformly  distributed   load  of 
intensity  w  is  fixed  at  A  and  merely  supported  at  B.     The  end  B  is 

w/4 

lowered  by  an  amount  -r-=-r.     Find  the  reactions.     How  much  must  B 
ioA/ 

be  lowered  so  that  the  whole  of  the  weight  may  be  borne  at  A? 

i  w/4 
Ans.  f|w/  at  A,  -&wl  at.  ^  ;  -  — — . 

8  £L1 


EXAMPLES. 

83.  Solve  the  preceding  question  supposing  the  fixture  at  A  to  be 
imperfect,  the  neutral  axis  making  with  the  horizontal  an  angle  whose 

tangent  is  — 0  -7^7.  Ans.  %wl,  \wl\  -~  -777. 

40  ±L1  4°  £'* 

84.  A  wrought-iron  girder  of  I-section,  2  ft.  deep,  with  flanges  of  equal 
area  and  having  their  joint  area  equal  to  that  of  the  web,  viz.,  48  sq.  in., 
carries  ^  ton  per  lineal  foot,  is  100  ft.  long,  consists  of  five  equal  spans, 
and  is  continuous  over  six  supports.     Find  the  reactions  when  the  third 
support  is  lowered  |-  in.     How  much  must  this  support  be  lowered  so 
that  the  reaction  may  be  nil  at  (a)  the  ist  support ;  (b)  the  3d  ;  (c)  the 
5th  ?     How  much  must  the  support  be  raised  so  that  the  reaction  may 
be  nil  at  (d}  the  2d  ;  (e)  the  4th  ;  and  (/)  the  6th  support  ?     E  =  16,500 
tons. 

Ans.  R^  =  2  If  ;     7?2  =  1 5f£  >  ^8  =  STF  » 

j?4  =  Hff  ;  Rs  =  9T*V ;  -#6  =  4T5¥  tons. 
(«)  if  in.;  mff  in.;  (c)  2ft  in. ; 
(d)  ijft  in. ;  00  2^  in. ;  (/)  6*  in. 

85.  If  the  three  supports  of  any  two  equal  consecutive  spans  of  a 
continuous  girder  of  any  number  of  spans  are   depressed  below  the 
horizontal,  show  that  the  relation  between  the  three  bending  moments 
at  the  supports  will  be  unaffected  if  the  depression  of  the  centre  support 
is  a  mean  between  the  depressions  of  the  other  two  supports. 

86.  A  girder  consists  of  two  spans  AB,  BC,  each  of  length  /,  and  is 
continuous  over  a  centre  pier  B.     A  uniform  load  of  length  2a  (<  /)  and 
of  intensity  W  travels  over  AB.     Find  the  reactions  at  the  supports  for 
any  given  position  of  the  load,  and  show  that  the  bending  moment  at 

aivl  f         a"*  \  i 

the  centre  pier  is  a  maximum  and  equal    to  --—.[-I  —  -s]    when  the 

3  1/3  V        rj 

centre  of  the  load  is  at  a  distance  I—       — j    from  A. 

87.  A  continuous  girder  rests  upon  three  supports  and  consists  of 
two  unequal  spans  AB  (=  /,),  BC  (=  /a).     A  uniform  load  of  intensity  iv 
travels  over  AB,  and  at  a  given  instant  covers  a  length  AD  (=  r)  of  the 
span.     If  Rlt  R3  are  the  reactions  at  A  and  C,  respectively,  show  that 

2    _  /       2  3  T      r* 

Draw  a  diagram  showing  the  shearing  force  in  front  of  the  moving 
load  as  it  crosses  the  girder. 

88.  If  the  live  load  in  the  preceding  question  may  cover  both  spans, 
show  that  the  shearing  force  at  any  point  D  is  a  maximum  when  AD 
and  BC  SLVQ  loaded  and  BD  unloaded. 

Illustrate  this  force  graphically,  taking  into  account  the  dead  load 
upon  the  girder. 


508  THEORY   OF  STRUCTURES. 

89.  A  continuous-girder  bridge  has  a  centre  span  of  300  ft.  and  two 
side  spans,  each  of  200  ft.  The  dead  load  upon  each  of  the  main  girders 
is  1250  Ibs.  per  lineal  foot.  In  one  of  the  side  spans  there  is  also  an 
additional  load  of  2500  Ibs.  per  lineal  foot  upon  each  girder.  Find  the 
reactions  and  points  of  inflexion.  How  much  must  the  third  support 
from  the  loaded  end  be  lowered  so  that  the  pressure  upon  it  may  be  just 
zero  ? 

Ans.  Let  W  =  weight  on  loaded  span  =  750,000  Ibs. 
R,  =  -ftfc  W  Ibs.  ;  Ri  =  |ff  I  W  Ibs.; 
R,  =  -/WV  ^  Ibs.  ;  Rt  =  tffr  W  Ibs. 
M,  =  —  \V/-  W  ft.-lbs.;  M*=—  ViV-  ^  ft.  -Ibs. 
Distance  of  point  of  inflexion  in  loaded  span  from  nearest 

end  support  =  i62ff  ft. 
Distance  of  point  of  inflexion  in  unloaded  end  span  from 

nearest  end  support  =  I45ff  ft. 

Distance  of  point  of  inflexion  in  intermediate  span  from 
end  support  in  unloaded  span  is  the  value  of  x  in  the 
equation  x*  —  £f  p,r  +  AJU^QJUL  _  o 

56350000   W 

3d  support  must  be  lowered  a  distance  =  -—  ^  --  . 

87        El 

90.  A  continuous  girder  AC  consists  of  two  equal  spans  AB,  BC,  each 
of  length  /,  and  carries  a  uniformly  distributed  load  of  intensity  w\  upon 
AB,  and  of  intensity  «/2  upon  BC.  Determine  the  bending  moments  at 
the  supports,  the  maximum  intermediate  bending  moments,  and  the  re- 
actions (a)  when  both  ends  of  the  girder  are  fixed  ;  (b)  when  one  end  A 
is  fixed  and  the  other  free. 

Ans.   Denoting  the  reactions  and  bending  moments  at  A,  B,  C 
by  Ri  ,  Mi  ,  y?2  ,  Mi  ,  Rs  ,  M3  ,  respectively  : 

r  /2 

(a)  Mi  =  —(—  50/1  +  wa)  ;    M  9  =  --  (wi  +  «/a); 

48  24 

r  RS 

M3  =  —  (wi  —  5^2);  Mmax.  in  AB  —  --  1-  Mi  ,  in  BC 

48  2Wi 

*  r> 

=  —  4- 

2W2  ID 

R*  =  —  (—o/i  +  90/2). 

ID 

/2  /a 

(b)  Mi  =  --  3(3^1  —  o/O;   Mi  =  —  ^-(o/i  +  20/2); 

28  28 

/?  a  /?  2 

M3  =  o  ;  Mmax.  mAB  =  —  +  Mlt   in  BC  = 

22£/i  20/ 


I20/a). 


EXAMPLES.  509 

91.   In  the  preceding  question,  if  Wi  =  iv*  =  w,  find  the  points  of  in- 
flexion and  the  maximum  deflection  in  each  case  and  for  each  span. 

Ans.  —  (a)  Points   of    inflexion   for  AB  or  BC   are  given   by 

6.r>-6;r/  +  /2  =  o. 
Max.  deflection  for  AB  or  BC  is  given  by 

—  Ely-  —  (2tx-x*-r), 
in  which  the  value  of  x  is  found  from 


(b)  Points  of  inflexion  in  AB  are  given  by 

I4.r2  —  i3.r/  +  2/2  =  o,   and  in  BC  by  x  = 
Max.  deflection  for  AB  is  given  by 


-  Ely  =  -==- 

»  I  DO 

and 

28.T2  —  39/JT  +    I2/2  =0. 

Max.  deflection  for  BC  is  given  by 


and 

28*3  —  33.^  +  4/3  =  o. 

92.  A  continuous  girder  ^4C  consists  of  two  equal  spans  AB,  BC  of 
15  m.  each.  Determine  the  bending  moments  at  the  supports,  the  maxi- 
mum intermediate  bending  moments,  and  the  reactions  (a)  when  the 
load  upon  each  span  is  3000  k.  per  metre  ;  (&)  when  the  load  per  metre  is 
3000  k.  upon  AB  and  1000  k.  upon  BC.  Call  M\  ,  M*,  M*  the  bending 
moments  and  R\  ,  Ri  ,  R3  the  reactions  at  A,  B,  C,  respectively,  and  con- 
sider three  cases,  viz.,  when  both  ends  of  the  girder  are  free,  when  both 
ends  are  fixed,  and  when  one  end  is  free  and  the  other  fixed. 
Ans.  —  Case  I  : 

(a)  M!  =  o  =  M9  ;  M*  =  —  84375  km.  ;  Mmax.  in  AB  or  BC 

—  47460.9375  km. 
fii  =  R3  =  I6875/&  ;  7?2  =  56250^. 
(&)  Mi  =  o  =  M3  ;    M*   =  —  56250  km.  ;    Mmax.   in  AB 

=  58593-75  km-»  in  BC  —  7031.25  km. 
Rl  =  18750  k.  ;  y?2  =  37500  k.;  7?8  =  3750  k. 
Case  II: 

(a)  Mi  =  M-!  =  M»  =  —  56250  km.;  Mmax.  in  AB  or  BC 
=  28125  km. 

r> 

7?i  =  -—  =  7?3  =  22500  k. 


510  THEORY  OF  STRUCTURES. 

(b)  Mi  —  —  65625  km.  ;  M*  =  —  37500  km.  ;  M3  =  —  9375 
km.  Mmax.  in  AB  =  33398.4375  km.,  in  BC 
=  6445.3125  km.; 

/?!   =  24375  k.;  R*  =  30000  k.;  R3  =  5625  k. 
Case  III  : 

(a)  Mi  =  —  48214!-  km.;  M9  =  —  72321$  km.;  Ms  =  o. 
Mma.v.  in  AB=  24537111  km.,  in  BC  =  52o88-j-f|  km.; 
R,   =  20892^  k.;  7?2  =  51428^  k.;  R3  =  482  if  k. 
(b}  Mi  =  —  64285!  km.;  Mi  =  —  401  78$  km.;  M,  =  o. 

Mmax.  in  AB  =  —  32573-^5-  km.,    in    BC  =  11623-^ 

km.  ; 
A'i   =  24107!  k.;  Ri  =  3  107  1  }  k.,  R3  =  4821!  k. 

93.  Show  that  a  uniformly  loaded  and  continuous  girder  of  two  equal 
spans,  with  both  ends  fixed,  is  2.08  times  as  stiff  as  if  the  ends  were  free 
and  merely  rested  on  the  supports. 

94.  A  single  weight  travels  over  the  span  AB  of  a  girder  of  two  equal 
spans,  AB,  BC,  continuous  over  a  centre  pier  B.     Show  that  the  reaction 

at  C  is  a  maximum  when  the  distance  of  the  weight  from  A  is   -~  if  the 

ends  A  and  C  rest  upon  their  supports,  and  when  the  distance  is  \AB  if 
the  two  ends  are  fixed.  Find  the  corresponding  bending  moments  at 
the  central  pier. 

PJ     "> 
Ans.  ---    -PL 


95.  A  girder  with  both  ends  fixed  carries  two  equal  loads  W  at  points 
dividing  the  girder  into  segments  a,  b,  c.  Determine  the  reactions  and 
bending  moments  at  the  supports. 

+  babe  +  3#V  +  2<r3  +  6a^  +  6bc* 
(a  +  b  +  c}3 
4- 


sty  —  yy  - 
If         ll7"* 

(a  +  b  +  cf 
c  +  2abc  4-  be*  +  ab^ 

M»         W2(ZC 

(a  +  b  +  r)2 
2  4-  2cibc  +  <%*b  +  b"*c 

(a  +  b  + 

96.  A  bridge  a  ft.  in  the  clear  is  formed  of  two  cantilevers  which. 
meet  in  the  centre  of  the  span  and  are  connected  by  a  bolt  capable  of 
transmitting  a  vertical  pressure  from  the  one  to  the  other.  A  weight 
W  is  placed  at  a  distance  b  from  one  of  the  abutments.  Find  the  pres- 
sure transmitted  from  one  cantilever  to  the  other,  and  draw  the  curve 
of  bending  moments  for  the  loaded  cantilever. 


Ans.    Ri=  W\  --,      + 

' 


;. 

2  --  ); 

a*) 


EXAMPLES.  5 1 1 

97.  The  weights  7,  7,  10,  10,  10,  10,  8,  8,  8,  8  tons,  taken  in  order  pass 
over  a  continuous  girder  of  two  spans,  each  of  50  ft.  and  fixed  at  both 
ends,  the  successive   intervals  being  5,  5,  5,  5,  5,  9,  5,  4,  5  ft.     Place  the 
wheels  so  as  to  give  the  maximum  bending  moment  at  the  centre  sup- 
port, and  find  its  value. 

Arts.   First  wheel  25.8399  ft.  from  nearest  abutment ; 
Max.  B.  M.  =  306.62  ft.-tons. 

98.  The  bridge  over  the  Grande  Baise  consists  of  two  equal  spans  of 
19.8  m. ;  each  of  the  main  girders  is  continuous  and  rests  upon  abut- 
ments at  the  ends.     Find  the   position  of  the   points  of   inflexion,  the 
bending   moment  at  the   centre   support,  the  maximum    intermediate 
bending  moment,  and  the  maximum  flange  stress  (a)  under    the  dead 
load  of  1700  k.  per  lineal  metre;  (b)  under  the  same  dead  load  together 
with  an  additional  proof  load  of  2000  k.  per  lineal    metre  on  one  span. 
The  depth  of  the  girder  =  3.228  m.,  and  /=  .093929232444. 

Ans. — (a)  14.85  m.from  the  abutments  ;  83308.5  kilogrammetres 

(km.);  46,861^  km.;  1.4315  k.  per  sq.  mm. 
(b)   16.18    m.  from  abutment   on  loaded  side;    11.876  m. 
from    abutment    on    unloaded    side;     132313.5  km.; 
101991.65625  km.;  2.27356  k.  per  sq.  mm. 

99.  The  Estressol  viaduct  consists  of  four  spans  of  25  m.  ;  the  main 
girders  are  continuous  and  their  ends  rest  upon  abutments ;  the  dead 
load    upon    each  girder  is  1700   k.    per   lineal    metre.     Determine   the 
position  of  points  of  inflexion  in  each  span,  the  reactions  and  bending 
moments  at  the  supports  when  an  additional  load  of  2000  k.  per  lineal 
metre  crosses  (a)  the  ist  span;   (b)  the   ist  and  2d  spans;    (<r)  all   the 
spans. 

Also,  find  the  absolute  maximum  bending   moments   at   the   inter- 
mediate supports. 

Ans.  Call  Xi ,  x*  x*,  x^  the  distances  of  points  of  inflexion  in 
ist,  2d,  3d,  and  4th  spans  from  the  ist,  2d,  4th,  and 
5th  supports,  respectively;  Ri ,  Ri,  R3,  R4,  R5  the 
reactions;  Mi  (=  o),  M<t,  M3,  M*,  Ms>  (=  o)  the 
bending  moments. 

(a)  XT.  =  20.72  m.  ; 

x*  is  given  by  I7oojr22  —  424017^2  +  395089!  =  o; 
xz  by  I7oo.r32  —  47767^3  +  238839!  =o  ;  jr4  =  19.38  m. 
l?i  =  38348TV  k.;  R*  =  81  i6of  k. ;  y?3  =  341071  k.; 
R,  =  499I0f  k. ;     ^  =  16473^  k. 
M-i—  197544^  km. ;  M3  =  5357if  km. ; 
km. 


$12  THEORY  OF  STRUCTURES. 

(b)   xl  =  19.556  m. ;  37ocxr22  —  103000^3  +  50357^  =  o; 

I700.T32  —  41071^3  +  205357^  =  o ;  XL  =  20. 168  m. 

7?!  =  361787  k. ;  ^2  =  107821^  k.;  fis  =  62964!  k.; 

/?4  =  45892^  k. ;  ^5  =  i7H2f  k. 

M*  =  258928^  km. ;  Ms  =  120535!-  km.; 

Mt  =  102678^  km. 

(e)   xi  =  19.64  m.  ;  -r2  and  jra  are  given  by 
i4^2-375.r-f  1875  =  0. 

Mt  —  Mi  =  247767%  km.  ;  Ma  =  165178^  km. 
Abs.  max.  B.  M.  at  2d  support  (=  max.  B.  M.  at  4th  sup- 
port) occurs  when  ist,  2d,  and  4th  spans  are   loaded, 

and  =  264508^1  km. 
Abs.  max.  B.  M.  at  3d   support   occurs  when  2d   and  3d 

spans  are  loaded  and  =  209821^  km. 

100.  In  the  preceding  question  find  the  absolute  maximum  flange 
unit  stress  at  the  piers,  /  being  .093929232444.     Ans.  4.5  k.  per  sq.  mm. 

101.  The  Osse   iron  viaduct  consists  of  seven  spans,  viz.,  two  end 
spans  of  28.8  m.  and  five  intermediate  spans  of  38  m. ;  each  main  girder 
is  continuous  and  carries  a  dead  load  of  1450  k.  per  lineal  metre.     Find 
the  bending  moments  at  the  supports  when  a  proof  load  of  2250  k.  per 
lineal  metre  for  each  girder  covers  all  the   spans;  and  also   find   the 
absolute   maximum    bending   moment   at  the  fourth  support.     Is   the 
following  section  of  sufficient  strength  ? — two  equal  flanges,  each  com- 
posed of  a  6oo-mm.  x  8-mm.  plate  riveted  by  means  of  two  loo-mm. 
x  ico-mm.  x  i2-mm.  angles  to  a  6oo-mm.  x  lo-mm.  vertical  web  plate 

and  two  8o-mm.  x  8o-mm.  x  ii-mm.  angles  riveted  to  each  horizontal 
plate  with  the  ends  of  the  horizontal  arms  15  mm.  from  the  edges  of 
the  plates;  the  whole  depth  of  the  section  being  4.016  m.,  and  the  dis- 
tance between  the  web  plates,  which  is  open,  being  2.8  m.  If  insuf- 
ficient, how  would  you  strengthen  it? 

Ans.  My  =  416,518  km.;  Ms  =  452,790  km.;  M^  =  443,722  km. 
Max.  B.  M.  =  542,199  km.     /=  .14074440467. 

.-.  —  =  .07009183, 

cM* 
and  max.  flange  stress  =  — —  =  7.73  k.  per  sq.  mm. 

This  is  much  too  large.  The  section  may  be  strengthened 
by  adding  two  6oo-mm.  x  8-mm.  plates  to  each  flange. 
/  is  thus  increased  by  .0783425536,  and  the  flange  unit 
stress  becomes  5  k.  per  sq.  mm. 


CHAPTER  VIII. 
PILLARS. 

1.  Classification. — The  manner  in  which  a  material  fails 
under  pressure  depends  not  merely  upon   its  nature  but  also 
upon  its  dimensions  and  form.     A  short  pillar,  e.g.,  a  cubical 
block,  will  bear  a  weight  that  will  almost  crush  it  into  powder, 
while  a  thin  plank  or  a  metal  coin  subjected  to  enormous  com- 
pression will  be  only  condensed  thereby.     In  designing  struts 
or  posts  for  bridges  and  other  structures,  it  must  be  borne  in 
mind  that  such  members  have  to  resist  buckling  and  bending  in 
addition  to  a  direct  pressure,  and  that  the  tendency  to  buckle 
or  bend  increases  with  the  ratio  of  the  length  of  a  pillar  to  its 
least  transverse  dimension. 

Hodgkinson,  guided  by  the  results  of  his  experiments, 
divided  all  pillars  with  truly  flat  and  firmly  bedded  ends  into 
three  classes,  viz. : 

(A)  Short  Pillars,  of  which  the  ratio  of  the  length  to   the 
diameter  is  less  than  4  or  5  ;  these  fail  under  a  direct  pressure. 

(B)  Medium  Pillars,  of  which  the  ratio  of  the  length  to  the 
diameter  exceeds  5,  and  is  less  than  30  if  of  cast-iron  or  tim- 
ber, and  less  than  60  if  of  wrought-iron  ;  these  fail  partly  by 
crushing  and  partly  by  flexure. 

(C)  Long  Pillars,  of  which  the  ratio  of  the  length   to   the 
diameter  exceeds   30   if   of  cast-iron   or  timber,  and   60  if   of 
wrought-iron  ;  these  fail  wholly  by  flexure. 

2.  Further    Deductions   from    Hodgkinson's    Experi- 
ments.— A  pillar  with  both   ends   rough  from  the  foundry  so 
that  a  load  can  be  applied  only  at  a  few  isolated  points,  and  a 
pillar  with  a  rounded  end  so  that  the  load  can  be  applied  only 

513 


514  THEORY  OF  STRUCTURES. 

along  the  axis,  are  each  one-tJiird  of  the  strength  of  a  pillar  of 
class  B,  and  from  one-third  to  two-thirds  of  the  strength  of  a 
pillar  of  class  C,  the  pillars  being  of  the  same  dimensions. 

The  strength  of  a  pillar  with  one  end  flat  and  the  other 
round  is  an  arithmetical  mean  between  the  strengths  of  two 
pillars  of  the  same  dimensions,  the  one  having  both  ends  flat 
and  the  other  both  ends  round. 

Disks  at  the  ends  of  pillars  only  slightly  increase  their 
strength,  but  facilitate  the  formation  of  connections. 

An  enlargement  of  the  middle  section  of  a  pillar  sometimes 
increases  its  strength  in  a  small  degree,  as  in  the  case  of  solid 
cast-iron  pillars  with  rounded  ends  which  are  made  stronger 
by  about  one-seventh ;  hollow  cast-iron  pillars  are  not  affected. 
The  strength  of  a  disk-ended  pillar  is  increased  by  about  one- 
eighth  or  one-ninth  when  the  middle  diameter  is  lengthened  by 
50  per  cent.,  but  for  slight  enlargements  the  increase  is  imper- 
ceptible. 

The  strength  of  hollow  cast-iron  pillars  is  not  affected  by  a 
slight  variation  in  the  thickness  of  the  metal,  as  a  thin  shell  is 
much  harder  than  a  thick  one.  The  excess  above  or  deficiency 
below  the  average  thickness  should  not  exceed  25  per  cent. 

3.  Form. — According  to  Hodgkinson,  the  relative  strengths 
of  long  cast-iron  pillars  of  equal  weight  and  length  may  be 
tabulated  as  follows  : 

(a)  Pillars  with  flat  ends. 

The  strength  of  a  solid  round  pillar  being  100, 

"  •"     square      "          is        93 ; 

"         "  triangular  "          is       1 10. 

(#)  Pillars  with  round  ends,  i.e.,  ends  for  hinging  or  pin 
connections. 

The  strength  of  a  hollow  cylindrical  pillar  being  100, 
"          "  an  H-shaped     .  "         is      74.6; 

"          "  a  +-shaped  "          is      44.2. 

The  strengths  of  a  long  solid  round  pillar  with  flat  ends, 
and  a  long  hollow  cylindrical  pillar  with  round  ends,  are  ap- 
proximately in  the  ratio  of  2.3  to  I. 

The  stiff cst  kind  of  wrought-iron  strut  is   a  built  tube,  the 


THE   FAILURE   OF  PILLARS.  515 

section  consisting  of  a  cell  or  of  cells,  which  may  be  circular, 
rectangular,  triangular,  or  of  any  convenient  form. 

In  experimenting  upon  hollow  tubes,  Hodgkinson  found 
that,  other  conditions  remaining  the  same,  the  circular  was  the 
strongest,  and  was  followed  in  order  of  strength  by  the  square 
in  four  compartments  j+j ;  the  rectangle  in  two  compartments, 
FT!  ;  the  rectangle,  a  ;  and  the  square. 

The  addition  of  a  diaphragm  across  the  middle  of  the  rect- 
angle doubled  its  resistance  to  crippling. 

4.  Modes  of  Failure. — The  manner  in  which  the  crush- 
ing of  short  pillars  takes  place  depends  upon  the  material,  and 
the  failure  may  be  due  to  splitting,  bulging,  or  buckling. 

(a)  Splitting  into  fragments  is  characteristic  of  such   crys- 
talline, fibrous,  or  granular  substances  as  glass,  timber, 
stone,  brick,  and  cast-iron. 

The  'compressive  strength  of  these  substances  is 
much  greater  than  their  tensile  strength,  and  when  they 
fail  they  do  so  suddenly. 

A   hard   vitreous   material,    e.g.,    glass    or   vitrified 
brick,  splits  into  a  number  of  prisms  (Fig.  335). 

A  fibrous  material,  e.g.,  timber,  and  granular  materials,  e.g., 
cast-iron   and  many  kinds  of   stone  and 
brick,  shear  or  slide  along  planes  oblique 
to  the  direction  of  the  thrust,  and  form 
one  or  more  wedges  or  pyramids  (Figs. 

336,  337,  238)- 

Sometimes  a  granular  or  a  crystalline  substance  will  sud- 
denly give  way  and  be  reduced  to  powder. 

(b)  Bulging,  i.e.,  a  lateral  spreading  out,  is  characteristic  of 
blocks  of  fibrous  materials,  e.g.,  wrought-iron,  copper,  lead,  and 
timber,  and  fracture  occurs  in  the  form  of  longitudinal  cracks. 

All  substances,  however,  even  the  most  crystalline,  will 
bulge  slightly  before  they  fail,  if  they  possess  some  degree  of 
toughness. 

(c)  Buckling  is  characteristic  of  fibrous   materials,  and  the 
resistance  of  a  pillar  to  buckling  is  always  less  than   its  resist- 
ance to  direct  crushing,  and  is  independent  of  length. 

Thin  malleable  plates  usually  fail  by  the  bending,  pucker- 


THEORY   OF  STRUCTURES. 

ing,  wrinkling,  or  crumpling  up  of  the  fibres,  and  the  same 
phenomena  may  be  observed  in  the  case  of  timber  and  of  long 
bars. 

Long  plate  tubes,  when  compressed  longitudinally,  first 
bend  and  eventually  fail  by  the  buckling  of  a  short  length  on 
the  concave  side. 

The  ultimate  resistance  to  buckling  of  a  well-made  and 
well-shaped  tube  is  about  27,000  Ibs.  per  square  inch  section  of 
metal,  which  may  be  increased  to  33,000  or  36,000  Ibs.  per 
square  inch  by  dividing  the  tube  into  two  or  more  compart- 
ments. 

A  rectangular  wrought-iron  or  steel  tube  offers  the  greatest 
resistance  to  buckling  when  the  mass  of  the  material  is  con- 
centrated at  the  angles,  while  the  sides  consist  of  thin  plates 
or  lattice-work  sufficiently  strong  to  prevent  the  bending  of 
the  angles. 

Timber  offers  about  twice  the  resistance  to  crushing  when 
dry  that  it  does  when  wet,  as  the  presence  of  moisture  dimin- 
ishes the  lateral  adhesion  of  the  fibres. 

5.  Uniform  Stress. — Let  a  short  pillar  be  subjected  to  a 
w|  pressure  of    W  Ibs.   uniformly  distributed  over   its 

end  and  acting  in  the  direction  of  its  axis. 

Let  5  be  the  transverse  sectional  area  of  the  pil- 
lar. 

W 

Let  /  =  — —  be   the  intensity  of  stress  per   unit 

O 

of  area  of  any  transverse  section  AB. 

TIG.  339.  J 

Let  A'B'  be  any  other  section  of  area   S',  in- 
clined to  the  axis  at  an  angle  0.     The   intensity  of  stress   per 

W       W 

unit  of  area  of  A'B'  =  —  =  ——  sin  6  =  p  sin  0,  which  may  be 
o  o 

resolved  into  a  component/  sin2  0  normal  to  A'B',  and  a  com- 
ponent/ sin  6  cos  6,  i.e.,/  -  — ,  parallel  to  A'B'.  The  last 

intensity  is  evidently  a  maximum  when  6  =  45°,  so  that  the 
plane  along  which  the  resistance  to  shearing  is  least,  and  there- 
fore along  which  the  fracture  of  a  homogeneous  material  would 
tend  to  take  place,  makes  an  angle  of  45°  with  the  axis. 


UNIFORM L  Y'VAR  YING   STRESS. 


517 


None  of  the  materials  of  construction  are  truly  homo- 
geneous, and  in  the  case  of  cast-iron  the  irregularity  of  the 
texture  and  the  hardness  of  the  skin  cause  the  angle  between 
the  plane  of  shear  and  the  direction  of  the  thrust  to  vary 
from  32°  to  42°.  Brick  chimneys  sometimes  fail  by  the  shear- 
ing of  the  mortar,  the  upper  portion  sliding  over  an  oblique 
plane. 

Hodgkinson's  experiments  upon  blocks  of  different  mate- 
rials led  him  to  infer  that  the  true  crushing  strength  of  a  ma- 
terial is  obtained  when  the  ratio  of  length  to  diameter  is  at 
least  i-J-;  for  a  less  ratio  the  resistance  to  compression  is  un- 
duly increased  by  the  friction  at  the  surfaces  between  which 
the  block  is  crushed. 

6.  Uniformly  Varying  Stress. — The  load  upon  a  pillar  is 
rarely,  if  ever,  uniformly  distrib- 
uted, but  it  is  practically  sufficient 
to  assume  that  the  pressure  in 
any  transverse  section  varies  uni- 
formly. 

Any  variable  external  force  ap- 
plied normally  to  a  plane  surface 
AA  of  area  5  may  be  graphically 
represented  by  a  cylinder  AABB, 
the  end  BB  being  the  locus  of  the 
extremities  of  ordinates  erected 
upon  A  A,  each  ordinate  being  pro- 
portional to  the  intensity  of  press- 
ure at  the  point  on  which  it  is 
erected. 

Let  P  be  the  total  force  upon  *AA,  and  let  the  line  of  its 
resultant  intersect  A  A  in  C\  C  is  the  centre  of  pressure  of  A  A, 
and  the  ordinate  CC  necessarily  passes  through  the  centre  of 
gravity  of  the  cylinder. 

Again,  the  resultant  internal  stress  developed  in  AA  is  P, 
and  may  of  course  be  graphically  represented  by  the  same 
cylinder  AABB. 

Assume  that  the  pressure  upon  AA  varies  uniformly;  the 
surface  BB  is  then  a  plane  inclined  at  a  certain  ano-le  to  AA. 


FIG.  340. 


5i8 


THEORY  OF  STRUCTURES. 


Take  0,  the  centre  of  figure  of  AA,  as  the  origin,  and  AA 
as  the  plane  of  x,  y. 

Let  O  y,  the  axis  of  y,  be  parallel  to 
that  line  EE  of  the  plane  BB  which  is 
parallel  to  the  plane  AA. 

Through  EE  draw  a  plane  DD  par- 
allel to  AA,  and  form  the  cylinder 
A  ADD. 

The  two  cylinders  A  ABB  and  A  ADD 
are  evidently  equal  in  volume,  and  OF, 
the  average  ordinate,  represents  the  mean 
pressure  over  AA  ;  let  it  be  denoted 

by  A- 

At    any    point   R  of  the  plane   AA, 
erect  the  ordinate  RQP,  intersecting  the 
planes   DD,  BB,  in   Q   and   P,    respect- 
Y/  ively. 

FlG-  341-  Let  x,  y  be  the  co-ordinates  of  R. 

The  pressure  at  R 


a  being  a  constant  depending  upon  the  variation. 

Note.  —  The  sign  of  x  is  negative  for  points  on  the  left  of  O,. 
and  the  pressure  at  a  point  corresponding  to  R\spQ  —  ax. 

Let  xn,  y0  be  the  co-ordinates  of  the  centre  of  pressure  C. 

Let  AS  be  an  elementary  area  at  any  point  R. 

Then  pAS  is  the  pressure  upon  AS,  and  2(J>4S)  is  the 
total  pressure  upon  the  surface  AA,  2  being  the  symbol  of 
summation. 

Hence, 


and 


But  /  =  /0  +  ax. 


and 


axy)AS\ 


UNIFORMLY    VARYING   STRESS. 


519 


Now  0  is  the  centre  of  figure  of  AA,  and  therefore  2 
and  ~2(ydS}  are  each  zero. 

Also,  2(4  S)  =  S,  2(x*4S)  is  the  moment  of  inertia  (/)  of 
A  A  with  respect  to  OY,  and  2Z(xydS)  is  the  product  of  inertia 
(K)  about  the  axis  OZ. 


.'.  x0p0S  =  al  —  x0P 


and 


(l) 
(2) 


Cor.  I.  Ip  any  symmetrical 
section  yQ  is  zero,  and  XQ  is  the 
deviation  of  the  centre  of  pres- 
sure C  from  the  centre  of  fig- 
ure  a 

Let  xl  be  the  distance  from 
O  of  the  extreme  points  A  of 
the  section. 

The  greatest  stress  in  A  A  is/0  +  axl  =  /\,  suppose. 


FlG"  342' 


•  •A-! 


or 


^> 
A 


It  is  generally  advisable,  especially  in  masonry  structures, 
to  limit  x0  by  the  condition  that  the  stress  shall  be  nowhere 
negative,  i.e.,  a  tension.  Now  the  minimum  stress  is/0  —  ax^  , 
so  that  to  fulfil  this  condition, 


pf  >  or  =  axl  ,     But  pl  =  ax, 


<  or  =  2/.. 


520  THEORY  OF  STRUCTURES. 

Hence,  by  eq.  (3), 

A,  i 


and  therefore 


I 

<  or  =  I  ;      i.e.,  x,  <  or  =     -. 


Cor.   2.    The  uniformly  varying  stress  is   equivalent  to  a 
single  force  P  along  the  axis,  and  a  couple  of  moment 


=  a  Vr  +  K\ 
Cor.  3.  The  line  CO  is  said  to  be  conjugate  to  OY. 

If  the  angle  COX  =  0,     then     cotV  =  —  =  ~. 

7.  Hodgkinson's  Formulae  for  the  Ultimate  Strength  of 
Long  and  Medium  Pillars.  —  When  a  Jong  pillar  is  subjected  to 
a  crushing  force  it  first  yields  sideways,  and  eventually  breaks 
in  a  manner  apparently  similar  to  the  fracture  of  a  beam  under 
a  transverse  load.  This  similarity,  however,  is  modified  by  the 
fact  that  an  initial  longitudinal  compression  is  induced  in  the 
pillar  by  the  superimposed  load. 

Hodgkinson  deduced,  experimentally,  that  the  strength  of 
long  solid  round  iron  and  square  timber  pillars,  with  flat  and 
firmly  bedded  ends,  is  given  by  an  expression  of  the  form 


W  being  the  breaking  weight  in  tons  of  2240  Ibs.  ; 
d        "        "    diameter  or  side  of  the  pillar  in  inches  ; 
I         "         "    length  of  the  pillar  in  feet; 
n  and  m  being  numerical  indices  ; 


FORMULAE  FOR    ULTIMATE   STRENGTH  OF  PILLARS.     $21 

A    being  a   constant  varying  with  the  material   and  with   the 
sectional  form  of  the  pillar. 

.  For  iron  pillars  ...............  n  =  3.6  and  m  =  1.7 

"  timber  pillars  ............  n  =  4     and  m  =  2 

"  cast-iron  ............................  A  =  44.16 

"  wrought-iron  ........................  A  =  133.75 

"  dry  Dantzic  oak  .....................  A  —  10.95 

"  dry  red  deal  .........................  A  —  7.81 

"  dry  French  oak  ......................  A  —  6*9 

The   strength  of  long  hollow  round   cast-iron   pillars  was 
found  to  be  given  by 


-  d™ 

w  -44.34  —  -r- 


d  being  the  external  and  dl  the  internal  diameter,  both  in 
inches. 

Thus,  the  strength  of  a  hollow  cast-iron  pillar  is  approxi- 
mately equal  to  the  difference  between  the  strengths  of  two 
solid  cast-iron  pillars  whose  diameters  are  equal  to  the  external 
and  internal  diameters  of  the  hollow  pillar. 

The  strength  of  medium  pillars  may  be  obtained  by  the 
formula 

w        Wfs 

- 


W  being  the  breaking  weight  in  tons  of  2240  Ibs.  ; 

W        "       "          "  "        "     "      "      "       "     as     derived 

from  the  formula  for  long  pillars  ; 

/being  the  ultimate  crushing  strength  in  tons  per  square  inch  ; 
S  being  the  sectional  area  of  the  pillar  in  square  inches. 

Again,  if  the  ends  of  a  cast-iron  pillar  are  rounded,  the 
above  formulae  may  be  still  employed  to  determine  its  strength, 
A  being  14.9  for  a  solid  and  13  for  a  hollow  pillar. 


522  THEORY  OF   STRUCTURES. 

8.  Gordon's  Formula  for  the  Ultimate  Strength  of  a 
Pillar.  —  The  method  discussed  in  the  preceding  articles,  being 
practically  very  inconvenient,  is  not  generally  used, 
and  the  present  article  will  treat  of  Professor  Gordon's 
formula,  which  has  a  better  theoretical  basis  and  is 
easier  of  application. 

The  effect  of  a  weight  W  upon  a  pillar  of  length  / 
;2/n     and  sectional  area  5  may  be  divided  into  two  parts  : 

(a)  A  direct  thrust,  which  produces  a  uniform  com- 

W  • 

pression  of  intensity  -=•  =  pl  . 

(b)  A  bending  moment,  which  causes  the  pillar  to 
FIG.  343.  yield  in  the  direction  of  its  least  dimension  (k). 

Letjy  be  the  greatest  deviation  of  the  pillar  from 
the  vertical. 

The  bending  moment  M  at  the  point  of  maximum  stress 
may  be  represented  by  Wy. 

Let  /2  be  the  stress  in  the  extreme  layers  due  to  this  bend- 
ing moment. 


Now 


c  being  the  distance  of  the  layer  under  consideration  from 
the  neutral  axis,  ju.  a  constant  depending  upon  the  sectional 
form,  and  b  the  dimension  perpendicular  to  the  plane  of  flexure. 


Wv 
and        « 


Butj/cc£.    (Art.  9,  Chap.  VI.)! 


wr     wr       /a 

'  and 


VALUES  OF  a  AND  /    GORDON'S  FORMULA.  $2$ 

a  being  some  constant  to  be  determined  by  experiment. 
Hence,  the  total  stress  in  the  most  strained  fibre  is 


or 


which  is  Gordon's  formula. 

Cor. — If  the  weight- upon  the  pillar  causes  the  stress  in  any 
transverse  section  to  vary  uniformly,  the  direct  thrust  in  the 

W\          *°2'S]  W 

extreme  layers  is  -~r\i  -| *— /   instead  of  -~,  (Cor.  i,  Art. 

»J  1  O 

6,)  ;FO  being  the  greatest  deviation  of   the   line   of   resultant 
thrust  from  the  axis  of  the  pillar. 

Let  k  be  the  radius  of  gyration  of  the  cross-section.     Then 


and  the  expression  for  the  direct  thrust  may  be  written 

W 


Hence,  Gordon's  formula  becomes 

/ 


- 

5     A 


9.  Values  of  a  and  /.  —  The   following   table,   giving   the 
values  of  the  constants  a  and/*  in  Gordon's  formula,  has  been 


524 


THEORY  OF  STRUCTURES. 


prepared  by  taking  an  average  of  the  best  known  results,  and 
is  applicable  to  round  and  square  pillars  with  square  ends. 


f  in  Ibs. 
per  sq.  in. 

a 

For  cast-iron  solid  rectangular  pillars  

80  ooo 

-r\r* 

"           "             "     round                 "     

80  ooo 

¥¥ff 
TJU 

"           "         hollow  rectangular    "     

80  ooo 

TOU 

*4» 

"           "               "       round             " 

80  ooo 

,ri* 

For  wrought-iron  solid  rectangular  pillars     .    . 

¥TTO 

-JU. 

"                   "             "     round                 "      

36  ooo 

^(TTJT 

«JL* 

"                   "         thick  hollow  round    "      

36  ooo 

*^M 

6?  2OO 

^FffT 

_JL- 

"     round                "     ', 

67  2OO 

^OtTTT 

*JU. 

67  2OO 

TTTTTT 
,  1,^. 

For  strong-steel  solid  rectangular  pillars  

1  14  ooo 

^?oiy 

«d&» 

"               "             "     round                 "     

1  14  ooo 

1400 
JL- 

"               "           hollow  round             "     

114  ooo 

^DTT 
i  . 

5  ooo 

T^ffTJ 
_1I 

"               "     round                "     

5  ooo 

^^^ 
_1_ 

7  2OO 

2s  ff 
_i 

¥^0 

If  Gordon's  formula  is  applied  to  pillars  with  pin  ends,  4*2 
takes  the  place  of  a ;  and  if  to  pillars  with  one  pin  end  and  one 
square  end,  |^z  takes  the  place  of  a. 

10.  Graphical  Comparison  of  the  Crushing  Unit  Strength 
of  Solid  Round  Cast-iron,  Wrought-iron,  and  Mild-steel 
Pillars. 

The  crushing  unit  stress  is  given  by/  = 


Take  the  different  values  of  j  as  abscissae,  and  the  corre- 
sponding values  of  /  as  ordinates ;  the  resulting  curves  are 
shown  in  Fig.  344. 

Hence,  the  strength  of  a  mild-steel  pillar  always  exceeds 
that  of  a  wrought-iron  pillar  but  is  less  than  that  of  a  cast-iron 

pillar  when  j  <  10.7  ;  a  wrought-iron  pillar  is  stronger  or  weaker 
than  a  cast-iron  pillar  according  as  r  >  or  <  28.5. 


APPLICATIONS   OF  GORDON'S  FORMULA. 


80,000  Ibs. 


28,486.4  Ibs. 


0 


in 


40         50        60 
FIG.  344. 


ii.  Application  of  Gordon's  Formula  to  Pillars  of  other 
Sectional  Forms. 

In  any  section  whatever,  the  least  transverse  dimension  for 
calculation  (i.e.,  ti)  is  to  be  measured  in  the  plane  of  greatest 
flexure. 

Thus,  it  may  be  taken  as  the  least  diameter  of  the  rectangle 
circumscribing  tee  (Fig.  345),  channel  (Fig.  346),  and  cruciform 
(Fig.  347)  sections,  and  as  the  perpendicular  from  the  angle  to  the 
opposite  side  of  a  triangle  circumscribing  angle(F\g.  348)  sections. 


FIG.  345. 


FIG.  346. 


FIG.  347. 


FIG.  348. 


From  a  series  of  experiments  upon  wrought-iron  pillars  of 
these  sections,  f  was  found  to  be  42,500  Ibs.,  and  a,  — . 

QOO 

In   cast-iron  struts  of  a  cruciform   section  /=  80,000  Ibs. 


and  a  —  —-. 

400 


526  THEORY   OF  STRUCTURES. 

These  results  are  only  approximately  true,  and  apply  to 
pillars  fixed  at  both  ends. 

12.  Rankine's  Modification  of  Gordon's  Formula.  —  The 
factor  a  in  Gordon's  formula  is  by  no  means  constant,  and  not 
only  varies  with  the  nature  of  the  material,  with  the  length  of 
the  pillar,  with  the  condition  of  its  ends,  etc.,  but  also  with  the 
sectional  form  of  the  pillar.  The  variation  due  to  this  latter 
cause  may  be  eliminated,  and  the  formula  rendered  somewhat 
more  exact,  by  introducing  the  least  radius  of  gyration  instead 
of  the  least  transverse  dimension. 

If  k  is  the  least  radius  of  gyration, 

/         mbh*  __  m 

—   —  r*  * 


—  -          /  T     — 

mass        non        n 

m  and  n  being  constants  which  depend  upon  the  sectional 
form.  Thus,  Gordon's  formula  for  pillars  with  square  ends 
may  be  written 


•  •.+:«,  j 

in  which  «,  is  independent  of  the  sectional  form,  all  variations 
of  the  latter  being  included  in  k*.  This  modified  form  of 
Gordon's  formula  was  first  suggested  by  Rankine. 

4al  is  substituted  for  al  if  the  pillar  has  two  pin  ends,  and 

Q 

-at  or  2#,  is  substituted  for  al  if  the  pillar  has  one  pin  end  and 

one  square  end. 
Rankine  gives 

for  wrought  iron,      f  •=.  36000  Ibs.,      -  =  36000; 
for  cast-iron,  /=  80000  Ibs.,      -  =  6400  ; 


for  dry  timber,  /=    7200  Ibs.,      -  ==  3000; 

a, 


KANKINE'S  MODIFICATION   OF   GORDON'S  FORMULA.     $2? 

In  good  American  practice  the  safe  working  unit  stress  in 
bridge  compression  members  is  determined  by  the  formula 


Safe  working  unit  stress  = 


f  being  8000  Ibs.  for  wrought-iron  and   10,000  Ibs.  for  steel, 
and  —  being  40,000  for  two  square  ends,  30,000  for  one  square 

and  one  pin  end,  and  20,000  for  two  pin  ends. 
Another  formula  often  employed  is, 

H\    .        /' 


/          Tf\ 
Working  stress  in  Ibs.  persq.  in.  X  (4  +  — )  == 


H  being  the  ratio  of  length  to  least  breadth,  where,  in  the  case 
of  wrought-iron, 

f  =  38,500  Ibs.  and  —  =  5820  for  two  square  ends; 


f  —  38,500     "      "   -  =  3000    "    one  square  and  one  pin  end. 


/'  =  37,800     "      "    -  =  1900    "    two  pin  ends. 

r_r 

^\\e  factor  of  safety,  viz.,  4  -| ,  increases  with  H,  and  par- 
tially provides  for  the  corresponding  decrease  in  the  strength 
to  resist  side  blows. 

EXAMPLES. — According  to  Rankine  the  ultimate  compres- 
sive  strength  of  wrought-iron  struts,  in  pounds  per  square 
inch,  is 

W  36000 


36000  £* 


528  THEORY   OF   STRUCTURES. 


If  the  section  is  a  solid  rectangle,  k?  =  —  ,  and  hence 

36000 


r 


1    3000  h* 

If  the  section  is  a  solid  circle,  /£2  =  -^,  and  hence 

36000 


A= 


7-.- 


1    2250  tf 

If  the  section  is  a  thin  annulus,  k*  =  -^  nearly,  and  hence 

36000 

1  +4500^ 


_If  -  is  small,    W=fS. 


If      is  large,   ^.      ,. 
Comparing  the  last  result  with  eq.  (5),  Case  4,  Art.  16, 


which  gives  a  theoretical  value  of  at  ,  the  actual  value  being 
somewhat  different. 

13.  Values  of  fca  for  Different  Sections. 

(a)  Solid  rectangle  :  k*  =  -~  =  —  ,  h  being  the  least  dimen- 

O  1  2 

sion. 

tx\    if  n  i       v       7         I  (bh*  -  b'h'*\ 

(b)  Hollow  rectangle:  k     =  -^  =  7^  bh  —  b'h'r     '  g 

the  greatest  and  least  outside  dimensions,  and  b',  h'  the  great- 
est and  least  inside  dimensions,  respectively. 


VALUES  OF  &*  FOR  DIFFERENT  SECTIONS.  $2$ 

Let  /  be  the  thickness  of  the  metal.     Then 

b'  =  b  —  2t     and     h'  =  h  —  2t, 
and  hence 


I    bh*-(b-2i)(h-2t}*         V 

~  12  bh  —  (b  —  2t)(h  —  2t)'     12  b  +  h> 

approximately,  when  t  is  small  compared  with  h,  i.e.,  for  a  thin 

hollow  rectangle. 

For  a  square  cell,  J?  =  -g-. 

(c)  Solid  triangle :  ft*  =  -~-  =  — =,  h  being  the  height. 

(d]  Hollow  triangle  :  1?  =  -^-  =  -—   , .  ^  ,7 . , ,  ^,  ^  being  the 
base  and  height  of  the  outside  triangle,  and  £',  h'  the  base  and 
height  of  the  inside  triangle,  respectively.     Also,  T>  =  77. 


_  ^  P  -  V*  i__tf  IP  +  bf*\ 

-k   ^i$F -&'*?-  is\    y    r 


Hence,  for  a  thin  triangular  cell,  &*  =  —  . 

r  ;r2 

(e)  Solid  cylinder  :  £*  =  —  =  —;,  k  being  the  diameter. 

(/)  Hollow  cylinder:  ff  =  ^  =  ~^(A'  +h'^  h  and  h'  being 
the  external  and  internal  diameters,  respectively. 

Hence,  for  a  thin  cylindrical  cell,  &  =  —  ,  approximately. 

EXAMPLE.  —  Gordon's  formula  for  hollow   cylindrical   cast- 
iron  pillars  is 

w__  _L  _  _  _  /_ 

~  "  a> 


. 

h  3 


500  /3  4000 


530  THEORY  OF  STRUCTURES. 


The  relation  /,  =  --  -»  may  be  assumed  to  hold  for 


hollow  square  struts  and  also  for  struts  of  a  cruciform  section. 

Ex.  i.  For  a  hollow  square  having  its  diagonal  equal   to 
the  internal  diameter  of  the  hollow  cylinder,  i.e.,  k  ', 


='  and  '•  = 


Ex.  2.  If  the  side  of  the  square  is  equal  to  the  external 
diameter,  i.e.,  h,  then 

k*  =  -z- ,     and    /,  = 


6  '  ^  ~  3     /'  ' 


(g)  Cruciform  section,    the    arms   being  equal: 


_ 

/  = h  — ;   S  —  zbk  —  ^ . 

12    '      12          12 


~  =    ' nearly- 


FIG.  349. 


Hence,  the  formula  for  a  cast-iron  pillar  of  cruciform  section 
may  be  written 


(/z)  Angle-iron  of  unequal  ribs,  the  greater  being  £  and  the 
less  A  : 


VALUES   OF  &*   FOR   DIFFERENT  SECTIONS. 


53  1 


ra 

Hence,  if  b  =  h,  i.e.,  if  the  ribs  are  equal,  ft?  —  — . 
(i)  Channel-iron,  the  dimensions  being  as  in  Fig.  350  : 

2bht\h  +  tY 


T  — 


12  ~  4(2At  +  6t) 

2ht  .        2bhf 


(  2t 

=  h\  71  +  4 

Also,         5  =  bt  +  2ht. 


=]  IF 


'  nearly' 


_     _         ___ 

1  2(2  At  +  bt)      4(2  At  +  bt}*  )  ' 


Let  the  area  of  the  two  flanges  —  A  =  2ht,  and  let  the 
area  of  the  web  =.  B  —  bt.     Then 


_    - 

12(A 


(fc)  \\-iron,  breadth  of  flanges  being  b,  length  of  web  h,  and 
thickness  of  metal  /  : 


73j  1    .3  73. 

/  =  2~  +  —  =  2--,  nearly  ;      5= 


At. 


•*'/P~  I22bt 


2bt 


I2A+B' 


A  being  the  area  of  the  flanges,  and  .5  the  area  of  the  web. 
(/)  Circular  segment,  of  radius  r  and  length  rO  : 


Hence,  for  a  semicircle,  since  0  =  ?r, 


532  THEORY  OF  STRUCTURES. 

(m)  Barlow  rail:    tf  =  -- ,  nearly. 

(n)  Two  Barlow  rails,  riveted  base  to  base:  £a  =  .393;-*, 
nearly. 

14.  American  Iron  Columns.— In  1880  Mr.  G.  Bouscaren 
read  before  the  American  Society  of  Civil  Engineers  a  paper 
containing  the  results  of  a  series  of  experiments  made  for  the 
Cincinnati  Southern  Railroad  upon  Keystone,  square,  Phoenix, 
and  American  Bridge  Co.'s  columns. 


EYSTONE  SQUARE  PHOENIX  '  AM.  BRIDGE  CO. 

FIG.  351.  FIG.  352.  FIG.  353.  FIG.  354. 

These  experiments  show,  as  those  of  Hodgkinson  and 
others  have  also  shown,  that  the  strength  of  iron  and  steel 
columns  is  not  only  dependent  on  the  ratio  of  length  to  diam- 
eter, and  on  the  forfh  of  the  cross-section,  but  also  on  the 
proportions  of  parts,  details  of  design  and  workmanship,  and 
on  the  quality  of  the  material  of  which  the  columns  are  con- 
structed. 

Further,  they  seem  to  lead  to  the  conclusions  that  Gordon's 
formula  is  more  correct  as  modified  by  Rankine,  and  that,  in 
the  case  of  columns  hinged  at  both  ends,  Rankine's  formula, 
with  tfj  assumed  at  double  the  value  it  has  when  the  formula  is 
applied  to  columns  with  flat  ends,  is  practically  correct. 

The  subjoined  table  gives  the  values  of  the  constants 
#,  and  /  as  deduced  from  Bouscaren's  experiments  by  Prof. 
W.  H.  Burr. 

In  1 88 1  Messrs.  Clarke,  Reeves  &  Co.  presented  to  the 
American  Society  of  Civil  Engineers  a  paper  containing  the 
results  of  experiments  upon  twenty  Phcenix  columns,  which 
appeared  to  show  that  neither  Gordon's  nor  Rankine's  formula 
expressed  the  true  strength  of  a  column  of  the  Phcenix  type. 
In  the  discussion  that  followed  the  reading  of  this  paper,  how- 
ever, it  was  demonstrated  that,  within  the  range  of  the  experi- 
ments, the  strength  of  intermediate  lengths  and  sections  of 


AMERICAN  IRON   COLUMNS. 


533 


/  in  Ibs. 


For  keystone  columns  with  flat  ends — swelled 

"  "  "  "       "       "    — straight      (open      or 

closed) 

"  "  "  "       "       "  —open  (swelled  straight) 

"       pin  ends— swelled 

For  square  columns  with  flat  ends 

"         "  "  "     pin  ends 

For  Phoenix  columns  with  flat  ends 

"  "  '*  "     round  ends 

"  "  "  "     pin  ends 

For  American  Bridge  Co.'s  columns   with  flat  ends 

"  "  "  "  "  "     round  ends 

"  "  4<  "  "  "     pin  ends 


36,000 

39>500 
38,300 
38,300 
39,000 
39.000 
42,000 
42,000 
42,000 
36,000 
36,000 
36,000 


18300 

1 
18300 

1 
12000 

1 
35000 

1 
17000 

1 
50000 

1 
12500 

1 
22700 

1 
46000 

1 
11500 

1 
21500 


Phoenix  columns  can  be  obtained  either  from  Rankine's  for- 
mula by  slightly  changing  the  constants,  or  from  very  simple 
new  formulae. 

Mr.  W.  G.  Bouscaren  showed  that  by  making  a.  = 


i  ooooo 


W 


and/=  38000,  the  calculated  values  of  -^-  agree   very  nearly 

o 

with  the  actual  experimental  results. 

Mr.  D.  J.  Whittemore  gave  the  following  (only  applicable 
for  lengths  varying  from  5  to  45  diameters)  as  expressing  the 
probable  ultimate  strength  of  these  columns  : 


Wlbs.  =  (1200  —  /T)3° 


H  being  the  ratio  of  length  to  diameter. 

Mr.  C.  E.  Emery  stated  that  the  ultimate  strength  in  each 
case  is  approximately  represented  by  the  formula 

M/1K         355Q63  +  3095Q# 
H+6.i7$    -' 

H  being  the  ratio  of  length  to  diameter. 


534  THEORY  OF  STRUCTURES. 

Taking  the  different  values  of  H  as  abscissae,  and  of  W  as 
ordinates,  this  is  the  equation  of  an  hyperbola.  It  agrees  very 
accurately  with  the  experimental  results  from  20  diameters 
upwards;  at  15  diameters  the  calculated  values  of  W  arc 
greater  than  those  given  by  the  experiments  ;  for  a  less  num- 
ber of  diameters  the  experimental  results  are  the  higher,  but 
the  variations  are  slight,  and  are  provided  for  in  the  factor  of 
safety. 

The  following  very  simple  formulae,  due  to  Prof.  W.  H. 
Burr,  give  results  agreeing  closely  with  those  obtained  in  the 
experiments : 

For  values  of  j  <  30,  the  ultimate  strength  in  pounds  per 
square  inch 

=  64700  —  4600.  IJL 
V  k' 

For  values  of  T  between  30  and  140,  the  ultimate  strength 
in  pounds  per  square  inch 

=  39640  — 46  j, 

k  being  the  radius  of  gyration. 

15.  Long  Thin  Pillar.— Let  ACB  be  the  bent  axis  of  a 
thin  pillar  of  length  /,  having  two  pin  ends  and  carry- 
ing a  load  W  at  B. 

Let  d  be  the  greatest  deviation  of  the  axis  from  the 
vertical.     Then 

E 
Wd  —  bending  moment  =  ^,-7,     .     .     (i) 

—  being  the  curvature  of  the  pillar  and  7  the  moment 
K 

v         of  inertia  of  the  most  strained  transverse  section. 
FJG.  355.        This  equation  is  only  true  on  the  assumptions  that— 

(1)  initially,  the  pillar  is  perfectly  straight ; 

(2)  initially,  the  line  of  action  of  the  load  coincides  with  the 
axis  of  the  pillar  ; 


LONG    THIN  PILLAR.  535 

(3)  the  material  of  the  pillar  is  homogeneous. 

These  assumptions  cannot  be  fulfilled  in  practice,  and  varia- 
tions from  theoretical  accuracy  may,  perhaps,  be  provided  for 
by  supposing  that  the  line  of  action  of  the  load  is  at  a  small 
distance  x  from  the  axis  of  the  pillar.  The  bending-moment 
equation  then  becomes 


f^  being  the  skin  stress  due  to  bending  at  a  distance  c  from  the 
neutral  axis. 

Again,  assuming  that  the  bent  axis  is  in  the  form  of  an  arc 
of  a  circle, 


f     ='/!,     ...    .    .    (4) 

and  consequently 

Wx 
d=-pI^W> '  •    (5) 

where 
'•  "'^    '  "•  '''  ''^"       P=^f- (Q 

If  the  line  of  action  of  the  load  W  coincided  with  the  axis 
of  the  pillar,  then  x  would  be  nil. 

Hence,  by  eq.  (5),  so  long  as  the  load  is  less  than  P,  d  —  o, 
and  the  failure  of  the  pillar  would  be  due  to  direct  crush- 
ing. If  the  load  is  equal  to  P,  d  would  become  indeterminate 

I—  — )  and  the  pillar  would  remain  in  a  state  of  neutral  equi- 
librium at  any  inclination  to  the  vertical. 

It  is  impossible  that  W  should  exceed  P,  as  d  would  then 
be  negative  ;  and  therefore  a  load  greater  than  P  would  cause 
the  pillar  to  bend  over  laterally  until  it  broke. 


536  THEORY  OF  STRUCTURES. 


Thus,  P  ~  —JT-  must  be  the  theoretical  maximum  compres- 

sive  strength  of  the  pillar. 

Again,  let  A  be  the  area  of  the  section  under  consideration  ; 
"    p  be  the  total  intensity  of  the  skin  stress  at  the 

section  ; 
"   f  be  the  intensity  of  the  direct  stress  due  to  W 

W_ 

~~  A  ; 

"    /",  be   the    intensity   of    the    stress    due    to  P 
P 


Then 


the  sign  of  f^  being  positive  for  the  compressed  side  of  the 
pillar  and  negative  for  the  side  in  tension. 


(8) 


k  being  the  radius  of  gyration. 

Let  h  be  the  least  transverse  dimension  of  the  section  in 
the  plane  of  flexure.     Then 

c  oc  h    and     k  also  a  h. 

c  _n 

*'•  fi  =  ~h  ' 

n  being  a  coefficient  depending  upon  the  form  of  the  section. 
For  a  rectangle,  n  =  6  ;  for  a  circle,  n  =  8  ;  also, 

Px 


LONG    THIN  PILLAR.  537 

Thus,  however  small  x  may  be,/  continually  increases  as 
the  difference  between  fl  and  f  diminishes.  The  pillar  will 
therefore  fail  for  some  value  of  /  less  than  the  theoretical 
maximum.  This  is  in  accordance  with  experience,  as  it  is 
found  that  a  small  load  causes  a  moderate  flexure  in  a  long 
pillar,  and  that  this  flexure  gradually  increases  with  the  load 
until  fracture  takes  place. 

In  no  case  should  /  exceed  the  elastic  limit,  as  in  such 
case  a  set  would  be  produced  and  the  deviation  x  would  be 
increased. 

If  the  tensile  strength  of  the  material  of  the  pillar  is  small, 
as  in  the  case  of  cast-iron,  failure  may  arise  from  the  tearing  of 
the  stretched  layers. 

Cor.  i.  The  above  also  applies  to  the  case  of  a  pillar  with 
one  end  fixed  and  the  other  free,  but  the  value  of  P  is 

2EI 
then  -^- . 

Cor.   2.    According   to    Euler  (see    following   article),    the 

7T2 

more  correct  value  of  P  is  pEIjt,  ^  being  I,  2,  J,  or  4,  accord- 
ing as  the  pillar  has  two  pin  ends,  one  fixed  end  and  one  end 
guided  in  the  direction  of  the  thrust,  one  fixed  and  one  free 
end,  or  two  fixed  ends. 

El  &  fh^ 

P  evidently  a  —^  a  EAj$  a  £-A\j 

Hence,  (a)  the  strength  of  a  long  pillar  is  proportional  to 
the  coefficient  of  elasticity ;  (b)  the  strengths  of  similar  pillars 
are  as  the  sectional  areas. 

Again,  ft  oc  -^  oc  d. 

But  Wd  =  --/  a  /2  a  d. 

Hence  W  is  approximately  constant,  and  the  weight  which 
produces  moderate  flexure  is  approximately  equal  to  the  break- 
ing weight. 

EXAMPLE. — Find  the  crushing  load  of  a  solid  mild-steel 
pillar  3  in.  in  diameter  and  10  ft.  long,  with  two  pin  ends. 


THEORY  OF  STRUCTURES. 

Also  find  the  deviation  (x)  of  the  line  of  action  of  a  load  of 
20,000  Ibs.  from  the  axis  of  the  pillar,  so  that  the  maximum 
intensity  of  stress  may  not  exceed  10,000  Ibs.  per  square  inch. 

By  Gordon's  formula  and  the  table,  page  524, 

the  crushing  load  =  --r- —    Viiw  =  85292.3  Ibs. 
~T  TTTFTV   -3    y 

Again,  the  theoretical  maximum  compressive  strength  P 
8  X  28000000   n(tf 

~5T     6l875lbs- 

/,  61875       99 


*  P-  w   /,-/•  41875     67* 

Hence 

20000  / 
10000  =  -3-3- 

or  x  =  .65  in. 

16.  Long  Columns  of  Uniform  Section.    (Ruler's  Theory.) 
CASE  I.   Columns  with  both  ends  hinged. 

f  — The    column  OA    of   length   /  is   bent 

[k  ,  under  a  thrust  P  and  takes   the    curved 

form  OMA. 

Take  O  as  the  origin,  the  vertical 
through  O  as  the  axis  of  x,  and  the  hori- 
zontal through  O  as  the  axis  of  y. 

Consider   a   section    at    any    point  M 
(x,y).     If  there  is  equilibrium  and  if  the 
line   of   action    of   P  coincides   with  the 
TT~"         axis  of  the  column,  the  equation  of  mo- 
FIG.  356.  ments  at  M  is 

;:;:;  . :      -EI&  =  M=**>    ::.::.,'_, 

or 

d'y  P 


LONG   COLUMNS   OF    UNIFORM   SECTION.  539 

dy 
Multiplying  each  side  of  the  equation  by  -      and  integrating, 


(2) 


b  being  a  constant  of  integration. 
dy  adx 

Integrating, 


sm-'.l-J  = 

or 

y  =  b  sin  (ax  +  c),      .     .     .     .     (3) 

c  being  a  constant  of  integration. 

When  x  =  o,  y  is  also  o,  and  hence  b  =  o  or  c  =  o. 

If  b  =  o,  y  is  always  o,  and  lateral  flexure  is  impossible. 

Take  c  =  o.     Then 

jj/  =  £  sin  ax (4) 

Also,  when  x  —  OA  =  OMA,  nearly,  =  /,  y  =  o. 

.*.  o  =  b  sin  al, 
or 


and  hence 

P=n*El"t (5) 


Now  the  /ft&rf  value  of  P  evidently  corresponds  to  n  =  ir 
and  hence  the  minimum  thrust  which  will  bend  the  column 
laterally  is 


540  THEORY   OF  STRUCTURES. 

Cor.  I.  If  the  column  is  made   to  pass  through  N  points 
dividing  the  vertical  OA  into  N  +  I  equal  divisions,  then 

y  =  o  when  x  = 

and  therefore,  by  eq.  (4), 

al 


or 

al 


and  hence 

i)3. 


FIG.  357. 

As  before,  the  least  value  of  P  corresponds  to  n  =  I,  and 


is  the  least  force  which  will  bend  the  column  laterally. 

Hence,  the  strength  of  the  column  is  increased  in  the  ratio 
of  4,  9,  16,  etc.,  by  causing  it  to  pass  through  points  which  divide 
its  length  into  2,  3,  4,  etc.,  equal  parts,  respectively. 

Cor.  2.  The  value  of  b  may  be  approximately  determined 
as  follows  :  \ 

Let  ds  =  length  of  element  at  M. 

Let  0  =  inclination  to  vertical  of  tangent  at  M. 

Then 

pressure  upon  ds  —  P  cos  6  =  P-r- 
and  the 


.     .,        S,        , 

compression  of  ds  =  ds  =  -=r-;dx, 

A  being  the  sectional  area  of  the  column. 

Hence,  the  total  diminution  of  the  length  of  the  column 


••!' 


.     7  /> 

:     /      — rf*__  /. 


LONG   COLUMNS   OF    UNIFORM   SECTION.  541 

Again,  the  length  of  the  column 

=f'(l  +  ^£^)dx  =  f(l  +  a'6'  cos'  ax}'dx' 

=   /    (  l  H  --  cos2  ax\dx,  approximately, 


Hence,  if  L  is  the  initial  length  of  the  column,  i.e.,  the 
length  before  compression, 


and  consequently 


EIL-l\          I 

b  =  2~ 


CASE  2.  Columns  with  one  end  fixed  and  the  other  constrained 
to  lie  in  the  same  vertical. 

Assume  that  the  lateral  deviation   is  prevented  ^ 
by  means  of  a  horizontal   force  H  at  the  top  of  a 
column.     Then 

TO  !       \ 

M 


-*)..     .     .     (I) 

MW 

A  particular  solution  of  this  is 

I 


,,  -  FIG.  358. 

^  y       d  u 


542  THEORY  OF  STRUCTURES. 

and  eq.  (i)  becomes 

<T* 

~  EId?  =  Pu> 
or 

d*u 

-&=-*«•       *,     *VV«      (2) 

The  solution  of  the  last  equation  is 

y  —  /  =  u  =  b  sin  (ax  +  c),  .     .    :,  '  "  v"  .     (3) 
b  and  c  being  constants  of  integration. 

TT 


(4) 


dy 
But  -—  =  o    when     x  =  o, 


and      j  =  o      when     ^r  =  o     and  when     x  —  /. 

rr 

.*  0=  —    ,  -\-abcosc\ 


TT 


O  =     sn 
Hence 

<2/-|-  ^  =  o    and     al  =  —  tan  c  =  tan  at, 

and  therefore 


=  4-493 


=  /V  £/' 


which  may  be  written  in  the  form 

£L 

/•• 


(5) 


LONG   COLUMNS   OF    UNIFORM   SECTION. 
It  is  sufficiently  approximate  to  write 

El 


543 


P=  27T5 


CASE  3.  Columns  with  one  end  fixed  and  the  other  free. 

A  rigid  arm  AB  is  connected  with  the 
free  end  A   of  a  column,  and  a  vertical  B 
force  P  applied  at  B  bends  the  column 
laterally,  until  its  axis  assumes  the  curved 
form  OMA. 

Let  AB  —  q,  A  C  =  p,  and  let  /  be  the 
length  of  the  column,  —  OC,  nearly. 

The  inclination  of  AB  to  the  horizon 
is  so  small  that  the  difference  in  length 
between  AB  and  its  horizontal  projection 
may  be  disregarded.  The  moment  equa- 
tion at  any  point  M  (x,  y)  is 


or 


(i) 


dy 
Multiplying  each  side  by  2—  and  integrating, 


b  being  a  constant  of  integration. 

dy 
But  -T-  =  o  when  y  =  o,  and  hence  b  =  o. 


(2) 


or 


544  THEORY  OF  STRUCTURES. 

Integrating, 


c  being  a  constant  of  integration,  or 

4)  4-  a  —  y 


But  y  =  o  when  x  =  o,  and  hence  c  =  o. 

.*. — — p =  cos«^.     .<,;.....   .    .     (4) 

Also,  y  ~  p  when  ^-  =  /. 

.*. — -: —  —  cos  a/. (5) 

If  q  is  very  small  or  nil,  the  term  ~rq~    may  be  disregarded, 

and  then 

o  —  cos  al. 


»  being  a  whole  odd  number. 

The  least  value  of  P  corresponds  to  n  =  i,  and  the  minimum 
pressure  which  will  cause  the  column  to  bend  laterally  is 


Cor.  I.  By  eq.  (5)  the  deviation  of  the  top  of  the  column 
from  the  vertical  is 

i  —  cos  al 


LONG   COLUMNS   OF   UNIFORM   SECTION.  54$ 

Cor.  2.  Let  the  force  applied  at  B  be  oblique  and  let  its. 
vertical  and  horizontal  components  be  P  and  //,  respectively. 
The  moment  equation  now  becomes 


A  particular  solution  of  this  is 

*)  .....    (10) 


Let  y  =  y1  -f-  u. 
Substituting  in  eq.  (9), 


or 


The  solution  of  this  equation  is 

u  =  b  sin  (ax  -f-  c)  —  y  —  y't 
b  and  c  being  constants  of  integration. 

TT 


(12) 


When  x  =  o,  /  and  ^  are  each  =  o ;  and  when  x  =  /,  y  =  q+ 

TT 

Hence,  o  =/  +  ?+  —  /-(-  ^  sin  ^; 

rr 

O  =  —  ^ -\- al>  cos  c ; 


546  THEORY  OF  STRUCTURES. 

three  equations  giving  b,  c,  and  p,  and  therefore  fully  deter- 
mining y. 

CASE  4.   Column  with  both  ends  fixed. 

Let  IJL  be  the  end  moment  of  fixture.     Then 


or 

-*>  +  «•*  =  «•(*-  JO,      .     .    (l) 


, 
where  b  =  -p. 


FW.  360.  Multiplying  each  side  of  the  equation  by  2^  and 

integrating, 


d  being  a  constant  of  integration. 

But  --  =  o  when  ^  —  o,  and  hence  d 


y*)  ......    (2) 

or 


I    Integrating, 


or 

-^  =  cos  (**  +  <:), (3) 


c  being  a  constant  of  integration. 


LONG  COLUMNS  OF   UNIFORM  SECTION.  547 

But  jj/  =  o  when  x  —  o  and  when  x  —  I.     Hence 

i  =  cos  c   and       I  =  cos  (al-\-  c). 
and  therefore     c  =  o          and     al  = 
n  being  a  whole  number.     Hence, 


or 


The  least  value  of  P  corresponds  to  n  =  I,  and  the  mini- 
mum thrust  which  will  cause  the  column  to  bend  laterally  is 


17.  Remarks.  —  From  the  preceding  it  appears  that  the 
maximum  theoretical  compressive  strength  of  a  column  per 
unit  of  area  may  be  expressed  in  the  form 


k  being  the  radius  of  gyration,  and  A  a  coefficient  whose  value 
is  i,  2,  J,  or  4,  according  as  the  column  has  two  hinged  ends, 
one  end  fixed  and  the  other  guided  in  the  direction  of  thrust,  one 
end  fixed  and  the  other  free,  or  two  fixed  ends. 

This  formula  is  easy  of  application,  but  Hodgkinson's 
experiments  show  that  the  value  of  P  as  derived  therefrom  is 
too  large.  This  may  be  partly  due  to  the  assumption  that  the 
elasticity  of  the  material  is  perfect. 

The  factors  of  safety  to  be  used  with  this  formula  vary 
from  4  to  8  for  iron  and  steel  and  from  4  to  15  for  timber. 

The  objection  to  the  use  of  flat  bars  as  compression  mem- 
bers has  sometimes  been  overestimated. 

Consider,  e.g.,  the  case  of  a  flat  bar  hinged  at  both  ends. 


THEORY  OF  STRUCTURES. 


Let  the  coefficient  of  elasticity  of  the  material  be  25,000,- 
ooo  Ibs. 

Let  the  working  stress  per  square  inch  be  8000  Ibs. 

The  bar  will  not  bend  laterally  under  pressure  so  long  as 


7T 

the  unit  stress  <  Etft,  and 


jrV  / 

8000  <  25000000-  7f ,     or     -7  <  50.7. 


Hence,  the  length  of  a  flat  bar  in  compression  seems  to  be 
comparatively  limited.  If,  however,  both  ends  are  securely 
fixed,  the  strength  is  quadrupled  and  the  admissible  length  of 
bar  is  doubled,  while  it  may  be  still  further  increased  by  fixing 
the  bar  at  intermediate  points  as  indicated  in  Corollary  i,  page 
540.  This  shows  the  marked  advantage  to  be  gained  by  rivet- 
ing together  the  diagonals  of  lattice-girders  at  the  points  where 
they  cross  each  other. 

p 

The  value  of  f  —.  —. r  (Art.  15)  must  not  exceed  the  elastic 
./i 

limit.  It  is  difficult  to  define  with  any  degree  of  accuracy  the 
elastic  limit  of  cast-iron  and  timber.  It  is  claimed,  indeed,  that 
the  latter  has  no  elastic  limit,  properly  so  called,  but  that  a 
permanent  set  is  produced  by  every  elastic  change  of  form.  It 
may  be  assumed,  however,  that  the  elasticity  of  these  materials 
is  practically  unaffected  so  long  as  they  are  not  loaded  to  more 
than  one  half  of  the  ultimate  crushing  load. 
Hence,  taking 

E  =  29,000,000  Ibs.  and  f  =  20,000  Ibs.  for  wrought-ironr 

£"  =  29,000,000   "  "  /=  33,600"      "    soft  steel, 

E  —  29,000,000   "  "  /  =  56,000   "       "    hard  steel, 

£=17,000,000   "  "  /  =  40,000   "      "    cast-iron, 

£=     1,500,000   "  "  /=    3,600   "      "    dry  timber, 


the  pillars  will  not  bend  laterally  unless  the  ratio  of  ->  or  — 

a        2r 


LONG  COLUMNS  OF   UNIFORM  SECTION.  549 

(d  being  the  shortest  side  of  a  rectangular  section  and  r  the 
radius  of  a  circular  section)  exceeds  the  values  given  in  the 
following  table : 


Material.                        Value  of -4-  Value  of —.  Formula. 

a  27" 

Wrought-iron 34.5  29.9 

Soft  steel 26.6  24.3 

Hard  steel 20.3  17.9      }•   f  =  '—  = 

Cast-iron 18.7  16.2 

Dry  timber 18.5  16 

Wrought-iron 48.8  42.3 

Soft  steel 37-7  34-4 

Hard  steel 28.8  25.3      V  ^  _  £  _ 

Cast-iron 26.4  22.9 

Dry  timber 26. i  22. 7 

Wrought-iron 17.2  14.9 

Soft  steel 13.3  12. i 

Hard  steel 10.1  8.9 

Cast-iron 9.3  8.1 

Dry  timber 9.2  8 

Wrought-iron 69  59.9 

Soft  steel 53.3  48.7 

Hard  steel 40.7  35.7      \  f  —  ^.  — 

Cast-iron 374  32.4 

Dry  timber 37  32 


Baker  has  deduced  by  experiment  the  following  formulae 
for  the  strength  of  wrought-iron  and  steel  pillars  of  from  10  to 
30  diameters  in  length  and  with  fixed  ends,  the  tensile  strength 
of  the  metals  ranging  from  20  to  60  tons  (2240  Ibs.)  per  square 
inch  : 

Let  t  be  the  tensile  strength  of  the  iron  or  steel,  and  H  the 
ratio  of  length  to  diameter. 

Then  the  ultimate  compressive  resistance,  in  pounds  per 
square  inch, 

for  solid  round  pillars  =  (.4    —  .oo6Ii)(t  +  18)  ; 

for  thin  tubes  =  (.44  —  .OO4//X/  +  18) ; 

for  tubes  with  stiffening  ribs  =  (.44  —  .OO2//)(V  4-  18) ; 
for  girder  sections  =  (.4      -  .oO4//)'/-f-  18). 


55O  THEORY  OF  STRUCTURES. 

18.  Weyrauch's  Theory  of  the  Resistance  to  Buckling. 

—  In  order  to  make  allowance  for  buckling,  Weyrauch  pro- 
poses the  two  following  methods  : 

METHOD  I.  Let  Fl  be  the  necessary  sectional  area,  and  £. 
the  admissible  unit  stress  for  a  strut  subjected  to  loads  vary- 
ing from  a  maximum  compression  Bl  to  a  minimum  com- 
pression By 

Let/7'  be  the  necessary  sectional  area,  and  b'  the  admissible 
unit  stress  for  a  strut  subjected  to  loads  which  vary  between  a 
given  maximum  tension  and  a  given  maximum  compression, 
B'  being  the  numerically  absolute  maximum  load,  and  B"  the 
maximum  load  of  the  opposite  kind. 

According  to  Art.  7,  Chap.  Ill,  if  there  is  no  tendency  to 
buckling, 

*  =  !=T-^IH  .....  c> 

1  4+-.f)     -  ,.  . 

and 


" 


If  there  is  a  tendency  to  buckling,  let  /  be  the  length  of 
the  strut,  F  its  required  sectional  area,  and  T  the  mean  unit 
.stress  at  the  moment  of  buckling. 

Then,  according  to  the  theory  of  long  struts, 


(3) 


d  being  a  coefficient  depending  upon  the  method  adopted  for 
securing  the  ends,  E  the  coefficient  of  elasticity,  and  /  the 
least  moment  of  inertia  of  the  section. 

Also,  let  t  be  the  statical  compressive  strength  of  the  ma- 
terial of  the  strut,  and  take  t  =  j*T.     Then 


RESISTANCE    TO  BUCKLING—  WE  YRA  UCH'S   THEORY.     55  1 


where  <r  =  —  —  ................     (5) 

If  the  strut  under  a  pressure  B  were  not  liable  to  buckling, 
it  would  be  subjected  to  a  direct  thrust  only.     The  required 

B 
sectional  area  of  the  strut  would  then  be  —  ,  and  the  unit  stress 

B 

for  an  area  F  would  be  -=;  . 
r 

If  the  strut  under  the  pressure  B  is  liable  to  buckling,  its 

ry 

required  sectional  area  will  be  •=  ,  since   T  is  the  mean   unit 

stress  at  the  moment  of  buckling.     Let  x  be  the  unit  stress  at 
the  moment  of  buckling,  for  the  area  F. 

Assuming  that  the  unit  stresses  in  the  two  cases  are  in  the 
same  ratio  as  the  required  sectional  areas,  then 


....       - 

*  ''  F  ::  T  ''  /' 


B   t          B 


The  force  which,  when  uniformly  distributed  over  the  area 
F,  will  produce  this  stress,  is  Fx  —  juB. 

Hence,  allowance  may  be  made  for  buckling  by  substitut- 
ing for  the  compressive  forces  in  equations  (i)  and  (2),  their 
values  multiplied  by  yu.  Thus,  equation  (i)  becomes 

'=  f  = 


and  equation  (2)  becomes 

F'  =  —  r—  =  —  -.  -      P"'  ^  B'  1S  a  comPressi°n> 


552  THEORY  OF  STRUCTURES. 

and 

n/  D/ 

F  —  ,,  =  —  —  -        ~~\>  if  B"  is  a  compression.         (9) 


If  /*  <  I,  equations  (i)  and  (2)  give  larger  sectional  areas 
than  equations  (7),  (8),  and  (9),  so  that  the  latter  are  to  be  ap- 
plied only  when  /*  >  I. 

METHOD  II.  General  formulae  applicable  to  all  values  of  /* 
may  be  obtained  by  following  the  same  line  of  reasoning  as 
that  adopted  in  the  proof  of  Gordon's  formula.  It  is  there 
assumed  that  the  total  unit  stress  in  the  most  strained  fibre  is 

72\ 

A  (l  +*!«/»  A  Demg  tne  stress  due  to  direct  compression,  and 

r 

p^a-j^  that  due  to  the  bending  action. 

So,  instead  of  employing  equations  (i)  and  (2)  when  ju  <  i, 
and  equations  (7),  (8),  and  (9)  when  ^  >  i,  formulae  including 
all  cases  may  be  obtained  by  substituting  for  the  compressive 
forces  in  equations  (i)  and  (2)  their  values  multiplied  by  I 

Thus,  equation  (i)  becomes 


and  equation  (2)  becomes 

F=  — 7—  — g?7 {,  if  B'  is  a  compression,    (n) 

V'(,_m' 


or 

B' 


comPressi°n- 


Equations  (7),  (8),  (9),  respectively,  give  larger  values  of  F 
than  the  corresponding  equations  (10),  (u),  and  (12). 


RESISTANCE    7V   BUCKLING—  IVEYRAUCH'  S    THEORY.    553 

Note.  —  For  wrought-iron  bars  it  may  be  assumed,  as  in  Arts. 
5,  6,  Chap.  Ill,  tha't  z/,  =  z/  ==  700  k.  per  sq.  cm.,  and  m^  =  m 

=1. 

The  value  of  <r  is  given  by  formula  (5),  but  is  unreliable,  and 
varies  in  practice  from  10,000  to  36,000  for  struts  with  fixed 
ends. 

When  the  ends  are  fixed,  8  —  47ra,  according  to  theory. 
Hence, 

o-  =  4*'  _. 

Therefore,  if  E  =.  2,000,000  k.  per  sq.  cm.,  and  t  —  3300  k. 
per  sq.  cm.,  $  —  23,926,  or  in  round  numbers  23,900  ;  24,000 
is  the  value  usually  adopted  by  Weyrauch. 

EXAMPLE.  —  The  load  upon  a  wrought-iron  column  360  cm. 
long  varies  between  a  compression  of  50,000  k.  and  a  compres- 
sion of  25,000  k.  Calculate  the  sectional  area  of  the  column, 
assuming  it  to  be  first  solid  and  second  hollow,  allowance  being 
made  for  buckling. 

First.  By  eq.  (i), 

50000  _  _400 
700(1  +ixttW)~'    7 
r  being  the  radius  of  the  section. 

Also,    /=—  . 
4 


r  50 

Hence,  by  eq.  (4), 

360  X  360       ii 
P  —  -  -  x  —  •  =  1.188. 

24000  50 

Thus  }JL  >  i,  and  by  eq.  (7)  the  required  sectional  area  is 
/^  X  1.188  =  AGO.  x  IsI88  _  57<9  sq>  cm< 

Second.     F,  =  Af&  =  ^(rtf  —  r22), 
r,  being  the  external  and  r2  the  internal  radius  of  the  section. 


554  THEORY  OF  STRUCTURES. 

Let  rl  —  9  cm.  and  r,  =  7.92  cm.     Then 
*(r?  ~  O  -  5743  sq.  cm. 


Also, 


'  I      r'  +  rf"  143-7264 
Hence,  by  eq.  (4), 

_  360X360  4  . 

24000          143,7264 

Thus,  in  the  latter  case,  since  ft  <  I,  there  is  no  tendency 
to  buckling. 

If  the  area  is  determined  by  equation  (10),  its  value  becomes 
1.15  X  ±$JL  =  65  sq.  cm. 

19.  Flexure  of  Columns.  —  In  Art.  16  the  moment  equa- 
tion has  been  expressed  in  the  form 


and  this  is  sufficiently  accurate  if  the  deviation  of  the  axis  of  the 
strut  from  the  vertical  is  so  small  that  [-T-]   may  be  neglected 

without  sensible  error. 

The  more  correct  equation  is 


p  being  the  radius  of  curvature. 

Consider,  e.g.,  the  strut  in  Art.  16,  Case  I.     Then 


P  i       dO      dO 

-j^-Ty  =  -  =  —  =  -j-  sin  6, 
Er        p       ds       dy 


FLEXURE   OF   COLUMNS.  555 

0  being  the  inclination  of  the  tangent  at  M  to  the  axis  of  *, 
and  ds  an  element  of  the  bent  strut  at  M. 
...  _  tfydy  —  sin  Odd. 
Integrating, 

,       .....     (i) 


0  being  the  value  of  0  at  a  strut  end. 

a  0 

Let  sin  —  =  p     and     sin  -  =  p  sin  0.     Then 


or 

y  =  —  cos  0 (2) 


^ 


Let  Y  be  the  maximum  deviation  of  the  axis  of  the  strut 
from  the  vertical,  i.e.,  the  value  of  y  when  0  =  o  or  0  =  o. 
Then 


2  sin  ~ 

--  2-  .........    (3) 

a 


Again, 

i  dB 

a  Vi  —  //  sin2  0 
Hence,  if  /be  the  length  of  the  strut, 

d<l>  2  , 


(4) 


.FM(0)  being  an  elliptic  integral  of  the  first  kind. 

Let  P  be  the  least  thrust  which  will  make  the  strut  bend. 
As  shown  in  Art.  16, 


556  THEORY  OF  STRUCTURES. 

and,  by  eq.  (4),  the  corresponding  value  of  the  modulus  yu  is 
given  by 

F^}=-2 (5) 

•  * 

Let  the  actual  thrust  on  the  strut  be 

P=n'P', .-..-    .     (6) 

H*  being  a  coefficient  >  unity. 

The  corresponding  value  of  the  modulus  is  given  by 


By  reference  to  Legendre's  Tables  it  is  found  that  a  large 

n 

increase  in  the  value  of  //,  i.e.,  of  sin  —  °  or  #0  ,  is  necessary  in 


order  to  produce  even  a  small  increase  in  the  value  of 

and  therefore  of  n{  —  A  /  —f  I.     Hence,  as  soon  as  the  thrust 


•(-V?) 


P  exceeds  the  least  thrust  which  will  bend  the  column,  viz., 
P',  00  rapidly  increases. 

The  total  maximum  intensity  of  stress  in  the  skin  of  the 
strut  at  the  most  deflected  point 

P       MB       P       PYz 


z  being  the  distance  of  the  skin  from  the  neutral  axis,  and  / 

p 
being  equal  to  —r. 

A 

The  last  term   of  this  equation  includes  the  product  fE, 

n 

which  is  very  large,  and  also  the  factor  sin  —  ,  which  increases 

with  #0  so  that  the  ultimate  strength  of  the  material  is  rapidly 
approached,  and,  in  fact,  rupture  usually  takes  place  before  the 
column  has  assumed  the  position  of  equilibrium  defined  by  the 
slope  #0  at  the  ends. 


FLEXURE   OF   COLUMNS.  557 

If  there  were  no  limit  to  the  flexure,  the  column  would 
take  its  position  of  equilibrium  only  after  a  number  of  oscilla- 
tions about  this  position,  and  the  maximum  stress  in  the 
material  would  be  necessarily  greater  than  that  given  by 
eq.  (8). 

Again, 

I  (i  —  2;*'  sin2  0X0 

dx  =  ds  cos  6  = . 

a     V I  —  p  sina  0 

Let  X  be  the  vertical  distance  between  the  strut  ends.    Then 


0 


0)  being  an  elliptic  integral  of  the  second  kind. 
Hence,  the  diminution  in  the  length  of  the  strut 


20.  Flexure  of  Columns  (Findlay).  —  In  a  paper  on  the 
flexure  of  columns  read  before  the  Canadian  Society  of  Civil 
Engineers  (Vol.  IV,  Part  I),  Findlay  expresses  the  moment 
equation  in  the  form 


p0  and  ——•  being  the  values  of  p  and  —  —  when  M  =  o. 


558  THEORY  OF  STRUCTURES. 

Hinged  Ends.  —  It  is  assumed  that  the  line  of  action  of  the 
thrust  P  is  at  a  distance  d  from  the  axis  of  the  strut.     Then 


'.    -  w 

or 


where  a*  =  -~j,  p  =  total  stress  at  the  distance  z  from  the 

neutral  axis,  and  /=  stress  due  to  direct  thrust  f  —  -jj,  so  that 

.the  stress  due  to  bending  =  p  —  f. 

It  is  also  assumed  that  the  form  of  the  axis  of  the  column 
before  it  is  acted  upon  by  the  thrust  P,  is  a  curve  of  sines 
defined  by  the  equation 

nx 
.y.  =  JcosT  ,....„•....     (4) 

the'origin  being  half-way  between  the  ends  of  the  strut,  and  A 
being  the  maximum  initial  deviation  of  the  axis  from  the  ver- 
tical, i.e.,  the  value  of  y^  when  x  =  o. 

d*y0            An*         nx 
.*.  —r-^  = j$-  cos  -y, 

and  hence,  by  eq.  (3), 

^2.-  V     4-d\-    A-  — 

A  solution  of  this  equation  is 

nx 

,  cos  ax   .      f  COS  T~ 


cos- 


FLEXURE   OF  COLUMNS.  $59 

Now  -  -  is  always  small  for  such  values  of  f  as  would  con- 
stitute a  safe  working  load,  and  therefore 

al               aT 
cos  -  —  i -£-,  approximately, 

so  that  eq.  (6)  becomes 


COS 


or 


-f  </  =  ^  cos  0*i  +  -yj  +  J  cos  -j-i  +  -^r,  approx.    (7) 

Let  Fbe  the  maximum  value  of  y,  i.e.,  the  value  of  y  when 
=  o.     Then 


Hence,  by  eq.  (3),  the  total  maximum  intensity  of  stress 

+4(rH.  (9) 


where     b  =  -  ( —  -I — ^  ]     and     c  =  - 

z 


Eq.  (9)  is  a  quadratic  from  which /may  be  found  in  terms 
of/.  As  a  first  approximation,/  may  be  substituted  for  f  in 
the  last  term  of  the  portion  within  brackets,  the  error  being  in 
the  direction  of  safety. 

Fixed  Ends. — Let  Ml  be  the  moment  of  fixture. 

Eq.  (3)  now  becomes 


560  THEORY  OF  STRUCTURES. 

Assuming  again  that  the  initial  form  of  the  axis  is  a  curve 
of  sines,  the  solution  of  the  last  equation  is 

nx 

cos  — 
MI       I  j  ,   Mf}  cos  ax  / 


cos- 


Initially, 

vn  =  A  cos 


and  -j-  is  equal  to  -7-°  when  x  =  —  or  = . 

dx  dx  22 

Hence, 

al 

sin  — 


al 

COS  —  I   — 


or 


Again,  the  value  of  y  at  the  point  x  =.  o  is 


Also,  if/j,  p9  are  the  total  maximum  intensities  of  stress  at 
the  end  and  at  the  most  deflected  point,  then 


— '    =  etc., 

ZJtL  \ 

and 

M\. 


two  equations  from  which  /  may  be  found  as  before. 

The  following  conclusions  are  drawn  from  the  above  inves- 
tigation : 

First.  The  actual  strength  of  a  column  depends  partly  upon 


FLEXURE    OF   COLUMNS.  56 1 

known  facts  as  to  dimensions,  material,  etc.,  and  partly  upon 
accidental  circumstances. 

Second.  Experiments  upon  the  crippling  or  destruction  of 
columns  cannot  be  expected  to  give  coherent  results  when 
applied  to  the  determination  of  the  constants  in  such  an  equa- 
tion as  No.  (9). 

Third.  It  is  a  question  whether  p  should  be  made  the 
elastic  limit  of  the  material  and  the  working  load  a  definite 
fraction  of  the  corresponding  value  of  f  derived  from  eq.  (9), 
or  whether/  should  be  the  allowable  skin  working  stress,  and 
the  working  stress /^  be  found  by  means  of  the  same  equation. 
The  former  seems  to  be  the  more  logical  assumption. 

Fourth.  It  would  appear  that  the  strength  of  hinged  col- 
umns is  likely  to  be  much  more  variable  than  the  strength  of. 
columns  with  fixed  ends,  as  it  depends  upon  two  variable 
elements  d  and  z/,  while  the  end  fixture  eliminates  d. 


Note. — The  Tables  on  the  following  page  give  the  numerical 
values  of  elliptic  integrals  of  the  first  and  second  kind,  and  are 
useful  in  applying  the  results  of  Art.  18. 


THEORY   OF  STRUCTURES. 
FIRST    ELLIPTIC    INTEGRAL, 


J> 

0° 

5° 
i<r 

15° 

20° 

o.ooo 

0.087 
0.175 
0.262 

0-349 

0.000 

0.087 

0.175 

0.262 

0.349 

o.ooo 
0.087 

0.175 

0.262 
0-349 

o.ooo 

0.087 

0.175 
0.262 

0.350 

0.000 

0.087 

0.175 
0.262 

0.350 

o.ooo 
0.087 

0.175 
0.263 

0.351 

o.ooo 
0.087 

0.175 
0.263 
0.352 

o.ooo 
0.087 

0.175 
0.263 

0-353 

o.ooo 
0.087 

0.175 

0.264 

0.354 

0.000 

0.087 

0.175 

0.264 

0-355 

0.000 

0.087 

0.175 

0.265 

0.356 

25 
30 

35 
40° 

45° 

0.436 
0.524 
0.611 
0.698 
0.785 

0.436 
0.524 

0.611 
0.699 
0.786 

0.437 
0.525 
0.612 
0.700 
0.789 

0.438 

0.526 
0.614 

0.703 
0.792 

0.439 
0.527 

0.617 

0.707 
0.798 

0.440 
0.529 
0.620 
0.712 
0.804 

0.441 

0.532 

0.624 

0.718 
0.814 

0-443 
0.536 
0.630 
0.727 
0.826 

0.445 
0.539 

0.636 

0.736 
0.839 

0.448 

0-544 
0.644 
0.748 
0.858 

0.451 
o  549 
0.653 
0.763 
0.88  1 

So° 

£ 

65° 
70° 

0.873 
0.960 
.047 
.134 

.222 

0.874 
0.961 
1.049 
•137 
.224- 

0.877 
0.965 
•054 
.143 
.232 

0.882 
0.972 
.062 

.153 
.244 

0.889 
0.981 

.074 
.168 
.262 

0.898 

0-993 
.090 

.187 
.285 

0.911 
.010 

.112 
.215 
•320 

0.928 

•034 
.142 
•254 
•370 

0.947 
I  060 
1.178 
1.302 
1.431 

0.974 
1.099 
1-233 
1.377 
1-534 

on 

.154 
.317 
.506 

.735 

75° 
80° 

85° 
90° 

.309 
.396 
.484 

•571 

.312 
.400 

.487 
•575 

.321 
.410 

.499 

.588 

.336 

.427 
.519 

.610 

•357 
•452 
.547 
.643 

.385 
•485 
.585 
.686 

.426 

•534 
.643 

.752 

.488 
.608 

•731 
1.854 

1.566 

1.705 
1.848 

1.993 

1.703 

1.885 
2.077  ' 
2.275 

2.028 

2.436 
3.131 

CO 

SECOND    ELLIPTIC    INTEGRAL. 


* 

It  =  0 

/*  =.i 

ft  =  .2 

V-  =  -3 

tt  —  .4 

f*  -  -5 

It  =  .6 

M  =  -7 

M  =  -8 

M  =-9 

M  =  I 

0° 

5° 
10° 

£ 

O.OOO 
0.087 

0.175 
0.262 

0-349 

0.000 

0.087 
0.175 

0.262 

0.349 

0.000 

0.087 
0.174 
0.262 
0.349 

O.OOO 

0.087 

0.174 

0.262 
0.348 

o.ooo 
0.087 

0.174 

0.261 

0.348 

o.ooo 
0.087 

0.174 

0.261 

0.347 

O.OOO 

0.087 
0.174 
0.261 
0-347 

O.OOO 
0.087 
0.174 
0.26O 
0.346 

O.OOO 

0.087 

0.174 

0.260 

0.345 

0.000 
0.087 
0.174 
0.259 
0-343 

O.OOO 
0.087 
0.174 
0.259 
0.342 

25° 
30° 

35° 
40° 

45° 

0.436 
0.524 

o.6n 
0.698 

0.785 

0.436 
0.523 

0.610 
0.698 

0.785 

0.436 
0.523 
0.609 
0.696 
0.782 

0.435 
0.521 
0.607 
0.693 
0.779 

0.434 

0.520 
0.605 

0.690 

0.773 

0.433 

0.518 

0.602 
0.685 
0.767 

0431 
0.515 
0.598 
0.679 

0-759 

0.430 
0.512 

0-593 
0.672 
0.748 

0.428 

0.509 
0.588 

0.664 

0.737 

0.425 
0.505 
0.581 
0.654 
0.723 

0.423 
O.5OO 

0-574 
0.643 
0./07 

i? 

60° 

65° 

70° 

0.873 
0.960 
.047 
•  134 

.222 

0.872 

o.959 
i.04b 
.132 
.219 

0.869 

0-955 
.041 
.126 

.212 

0.864 
0.948 
1.032 
.116 
.200 

0857 
0.939 

.021 
.103 
.184 

0.848 
0.928 
.008 

.086 
.163 

0.837 
0.914 
0.989 
.063 
.135 

0.823 
0.895 
0.965 

1.033 
I.0<)9 

0.808 
0.875 
0.940 

I.OOI 

1.  060 

0.789 
0.850 
0.007 
O.QdO 
I.OOS 

0.766 
0.819 

0.866 
0.906 
0.940 

g: 
£ 

.309 
.396 

.484 

I-57I 

.306 

-393 
.480 
.566 

.297 
.383 
.468 

1-554 

-283 
.367 
•450 
•533 

.264 

•344 

.424 

•504 

.240 

.316 
•392 
.467 

.207 

•  277 
•347 
.417 

I.l"63 

1.227 
1.289 
I-35I 

1.117 
1.172 

1.225 
1.278 

1-053 

1.095 

I-I35 
I.I73 

0.966 
0.985 
0.996 

I.OOO 

EXAMPLES.  563 


EXAMPLES. 

1.  A  Phoenix  column  in  four  segments,  each  weighing  17  Ibs.  per 
lineal  yard,  carries  a  load  of  68,000  Ibs.     What  is  the  compressive  unit 
stress  ?  Ans.  10,000  Ibs.  per  sq.  in. 

2.  The  sectional  area  of  a  pillar  is  144  sq.   in.,  and  the  pillar  carries 
a  load  of  4000  Ibs.     Find  the  normal  and  tangential  intensities  of  stress 
on  a  plane  inclined  at  20°  to  the  axis.  Ans.  3.25  Ibs.;  8.93  Ibs. 

3.  A  short  hollow  square  column  has  to  support  a  load  of  120,000  Ibs., 
the  allowable  stress  being  15,000  Ibs.  per  square  inch.     Find  the  thick- 
ness of  the  metal,  an  external  side  of  the  column  being  6  in. 

Ans.  .36  in. 

4.  A  solid  cast-iron  pillar  9  ft.  in  height  and  4  in.  in  diameter  sup- 
ports a  load  of  55,000  Ibs.     Find  the  normal  and  shearing  intensity  of 
stress  per  square  inch  in  a  plane  section  inclined  at  30°  to  the  axis. 

If  the  ends  of  the  pillar  are  flat  and  firmly  bedded,  determine  its 
breaking  weight,  both  by  Hodgkinson's  and  by  Gordon's  formula. 

Ans.  1093!  Ibs. ;  1894.375  Ibs. ;  141^  tons  by  H.;  159  tons  by  G. 

5.  A  cylindrical  pillar  6  in.  in  diameter  supports  a  load  of  400  Ibs., 
of  which  the  centre  of  gravity  is  $•.  in.  from  the  axis.     Determine  the 
greatest  and  least  intensities  of  stress  upon  any  transverse  section  of 
the  pillar.  Ans.  25$$  Ibs. ;  2jjff  Ibs. 

6.  Compare  the  breaking  weights  of  round  cast-iron,  wrought-iron, 
and  mild-steel  pillars  with  flat  and  firmly  bedded  ends,  each  being  9  ft. 
in  length  and  6*  in.  in  diameter. 

Ans.  1,250,197  Ibs. ;  890,109  Ibs. ;  1,543,572  Ibs. 

7.  A  hollow  cast-iron  pillar  with  an  external  diameter  of  9  in.  is  to 
be  substituted  for  the  solid  pillar  in  the  preceding  question.     Determine 
the  thickness  of  the  metal.  Ans.  f  in. 

8.  Determine  the  breaking  weight  of  a  solid  round  pillar  with  both 
ends  firmly  secured,   10  ft.  in   length  and  2  in.   in  diameter,  (r)  if  of 
cast-iron  ;   (2)  if  of  wrought-iron  ;  (3)  if  of  steel  (mild). 

Ans.  25142.8  Ibs.;  43516.48105.;  54^36  Ibs. 

9.  A  hollow  cast-iron  pillar  12  ft.  in  height  has  to  support  a  steady 
load  of  33,000  Ibs.;  its  internal  diameter  is  5^  in.     Find  the  thickness 
of  the  metal,  the  factor  of  safety  being  6.  Ans.   .28  in. 

10.  A  solid  wrought-iron  pillar  is  to  be  substituted  for  the  pillar  in 
the  preceding  question.     Find  its  diameter.  Ans.  3^  in. 


564  THEORY  OF   STRUCTURES. 

11.  What  is  the  breaking  weight  of  a  hollow  cast-iron  pillar  9  ft.  in 
length  and  6  in.  square,  the  metal  being  i  in.  thick  ? 

Ans.  970873.6  Ibs. 

12.  Compare  the  breaking  weight  of  a  solid  square  pillar  of  wrought- 
iron  20  ft.  long  and  6  in.  square  with  that  of  a  solid  rectangular  pillar 
of  the  same  material,  the  section  being  9  in.  by  4  in. 

Ans.  845,217  Ibs.;  589,090  Ibs. 

13.  Compare  the  breaking  weights,  as  derived  from  Hodgkinson's  and 
Gordon's  formulae,  of  a  solid   round  cast-iron  pillar  20  ft.  in  length  and 
10  in.  in  diameter,   (i)  both  ends  being  securely  fixed ;  (2)  both  ends 
being  imperfectly  fixed. 

Ans. — (i)  951.4  tons  by  H.;  1150.05  tons  by  G. 
(2)  280.6  tons  by  H.;  415  tons  by  G. 

14.  Determine  by  Hodgkinson's  formula  the  diameter   of   a  solid 
wrought-iron  pillar  equal  in  length  and  strength  to  that  in  the  preceding 
question.  Ans.  7.35  in. 

15.  A  solid  or  hollow  pillar  of  cast-iron,  wrought-iron,  or  mild  steel 
is  to  be  designed  to  carry  a  steady  load  of  30,000  Ibs.     Determine  the 
necessary  diameter  in  each  case,  6  being  a  factor  of  safety.     (The  pillar 
is  to  be   12  ft.  high,  and  the  metal  of  the  hollow  pillar  is  to  be  f  in, 
thick.)  Ans. — Solid:       3.42^.;  3.25  in.;  2.8  in. 

Hollow:  4. 5'  in.;  4.75  in.;  3.5  in. 

16.  Determine  the  load  in  the  preceding  question  that  will  produce 
a  maximum  stress  of  9000  Ibs.  per  square  inch  in  the  solid  steel  pillar. 

17.  A  pillar  of  diameter  D  supports  a  given  load  ;  if  N  pillars,  each 
of  diameter  d,  are  substituted  for  this  single  pillar,  show  that  d  must  lie 

D        ,    D 

between  — -  and  — -. 
TV*          N* 

Is  it  more  economical  to  employ  few  or  many  pillars  of  given  height 
to  support  a  given  load? 

1 8.  A  solid  round  pillar  of  mild  steel,  16  ft.  high,  supports  a  steady 
load  of  20,000  Ibs.     If  the  factor  of  safety  is  6,  what  is  its  diameter  ? 

Ans.  3  in. 

19.  Find  the  diameter  of  each  of  four  pillars  of  the  same  material 
which  may  be  substituted  for  the  single  pillar  in  the  preceding  question. 

Ans.   2.04  in. 

20.  What  is  the  breaking  weight  of  a  cast-iron  stanchion  of  a  regular 
cruciform  section  and  15  ft.  in  height,  the  arms  being  24  in.  by  i  in.  ? 

Ans.  2,811,215  Ibs. 

21.  Each  of  the  pillars  supporting  the  lowest  floor  of  a  refinery  is 
6$  ft.  high,  is  of  a  regular  cruciform  section,  and  carries  a  load   of 


EXAMPLES.  565 

240,000  Ibs. ;  the  total  length  of  an  arm  is  26  in.     Determine  its  thick- 
ness, the  factor  of  safety  being  10.  Ans.  2.558  in. 

22.  Find  the  breaking  stress  per  square  inch  of  a  4-in.  x  4-in.  solid 
wrought-iron  pillar  for  lengths  of  5,  10,  15,  and   20  ft.,  the  two  ends 
being  absolutely  fixed. 

Ans.  33,488  Ibs.;  27,692  Ibs.;  21,492  Ibs.;   16,363  Ibs. 

23.  Find  the  diameter  of  a  wooden  column  20  ft.  long,  to  support  a 
load  of   10,000  Ibs.,   10  being  a  factor  of  safety  and  both  ends  of  the 
column  being  absolutely  fixed.  Ans.  8.55  in. 

24.  The  external  and  internal  diameters  of  a  hollow  cast-iron  column 
12  ft.  in  length  are  D  and  |Z?(  respectively;  the  load  upon  the  column 
is  25,000  Ibs.      If  the  factor  of  safety  is  4,  find  D,  (a)  when  both  ends  of 
the  column  are  absolutely  fixed ;   (£)  when  both  ends  are  hinged. 

Ans.  (a)  4.3  in.;  (b)  5.4  in. 

25.  Determine  the  breaking  weight  of  an  oak  pillar  9  ft.  high,  n  in. 
wide,  and  5  in.  thick.  Ans.   138,160  Ibs. 

26.  What  weight  will  be  safely  borne  by  a  pillar  of  dry  oak  subject 
to  vibration,  10  ft.  high  and  6  in.  square,  10  being  a  factor  of  safety? 

Ans.  9969  Ibs. 

27.  The  web  members  of  a  Warren  girder  are  bars  of  rectangular 
section  and  10  ft.  in  length.     One  of  the  bars  has  to  carry  loads  varying 
between  a  steady  maximum  tension  of  20.2  tons  and  a  maximum  tension 
of  40.4  tons,  and  another  to  carry  loads  varying  between  a  maximum 
compression  of  8.7  tons  and  a  maximum  tension  of  14.4  tons.     Find  the 
sectional  area  in  each  case,  allowance  being  made  for  buckling  in  the 
latter. 

28.  Determine  the  sectional  area  of  a  double-tee  strut  which  is  to 
carry  a  load  varying  between  a  maximum  tension  of  80,000  Ibs.  and  a 
maximum  compression  of  60,000  Ibs.     Each  flange  consists  of  two  6- 
in.  x  6-in.  x  f-in.    angle-irons    riveted    to    a    i2-in.  x  f  in.    web  plate. 
The  length  of  the  strut  is  to  be  (a)  6  ft.;  (8)  12  ft. 

29.  A  steel  strut  10  ft.  long  consists  of  two  tees  back  to  back,  each 
4 in.  X4in.  x  ^in.     Taking/  =  60,000  Ibs.,  ai  =  77^VT  (page  526),  and  6 
as  a  factor  of  safety,  find  the  working  load  (a)  when  the  strut  has  two 
pin  ends;  (b)  when  it  has  two  fixed  ends.     (E  =  29,000,000  Ibs.) 

Also,  find  the  deviation  of  the  axis  of  the  load  from  the  axis  of  the 
strut  so  that  the  maximum  stress  in  the  metal  may  not.  exceed  10,000 
Ibs.  per  square  inch. 

Ans —(a)  25, 585  Ibs. ;  (ff)  52,229  Ibs. 

Deviation  =  .55  in.  in  (a}  and  .158  in.  in  (&). 

30.  A  solid  wrought-iron  strut  20  ft.  high  and  4  in.  in  diameter  has 
one  end  fixed  and  the  other  perfectly  free.     Find  the  deviation  of  the 


566  THEORY  OF  STRUCTURES. 

line  cf  action  of  a  load  of  10,000  Ibs.  from  the  axis,  so  that  the  stress 
may  not  exceed  10,000  Ibs.  per  square  inch,  E  being  27,000,000  Ibs. 

A?is.  .88  in.  if  P  =  ^— -  ;      1.8  in.  if  P  —  -EI^. 

31.  A  hollow  cast-iron  column  with  two  pin  ends  is  24  ft.  high  and 
has  a  mean  diameter  of   12  in. ;  it  carries  a  load  of  80,000  Ibs.     Find  the 
proper  thickness  of  the  metal,    10  being   a  factor  of   safety.      If  the 
deviation  of  the  line  of  action  of  the  load  from  the  axis  is  i  in.,  find 
the  maximum  stress  per  square   inch  in  the  metal,  E  being   17,000,000 
Ibs.  Ans.  1.28  in.;  2236  Ibs.  per  sq.  in. 

32.  Find  the  crushing  load  of  a  solid  wrought-iron  pillar  3  in.  in 
diameter,  10  ft.  high,  and  fixed  at  both  ends.     Calculate  the  deviation 
which  will  produce  a  maximum  stress  in  the  metal  of  9000  Ibs.  per 
square   inch  under  loads  of   (a)    15,000  Ibs.;  (£)    30,000   Ibs.;  E  being 
29,000,000  Ibs.  Ans.   148,775  Ibs.     (a)  1.158  in.  ;  (£)  .38  in. 

33.  Solve  the  preceding  question  on  the  assumption  that  the  column 
has  two  pin  ends.  Ans.  66,218  Ibs.  ;  (a)  .985  in.;  (b)  .261  in. 

34.  A  pier  consists  of  N  rows  of  posts  equidistant  from  each  other, 
N  being  even;  d  is  the  distance  from  centre  to  centre  of  the  outside 
rows;    W  is  the  gross  vertical  load  upon  the  pier;  H  is   the  greatest 
horizontal  thrust,  and  acts  upon  the  pier  at  a  height  y  above  the  base. 
Assuming  the  principle  of  a  uniformly  varying  stress,  the  portion  of  the 
load  borne  by  the  n-th  row  of  posts  measured  from  the  centre  line  is 

1 . — .     Find  the  value  of  the  coefficient  a  in  terms  of  d,  H, 

N         -2  N  —  \ 

y,  and  N,  and  determine  the  best  value  for  d. 

35.  Prove  that  the  flexural  rigidity  of  a  straight  beam,  of  sectional 

area  A,  under  a  thrust  P  per  unit  of  area,  is  EAk*l  i  —  — ).  and  that  the 
beam  will  bend  if  its  length,  when  unstrained,  exceeds 


Ak*  being  the  moment  of  inertia  of  the  section,  and  E  the  coefficient  of 
elasticity  of  the  material. 

36.  Find  the  safe  load  on  a  rolled  tee-iron  strut  6  in.  x  4  in.  x  %  in., 
10  ft.  long,  fixed  at  one  end,  free  at  the  other. 

37.  In  Art.  19,  show  how  equations  (3)  and  (6)  will  be  modified  if  the 
line  of  action  of  P  is  distant  a  +  ft  from  one  end  and  a  —  ft  from  the 
other  end  of  the  column's  axis.     Also,  if  the  coefficient  of  elasticity,  E, 

is  variable  and  equal  to  in  ±  n-  at  a  point  distant  z  from  the  axis,  r  being 


EXAMPLES.  567 

the   .maximum    value    of    z,   and    m    and     n    coefficients,   show    that 

n  k" 
y  +  -     —  must  be  substituted  for y  in  eq.  (3). 

38.  In  one  of  Christie's  experiments  an  angle-bar  2  in.  x  2  in.  x  -fa  in., 
with  hinged  ends,  for  which  —  had  the  value  154,  deflected  .01  in.  for  an 

increase  in  the  load  of  3000  Ibs.     Show  that  -Q  H — 9  =  .01  in. 

o          n 

39.  A  long  column  with  pin  ends  is  bent  laterally  until  the  angular 
deviation  (00)  at  the  ends  is  4°.     Find  the  total  maximum   intensity  of 
stress,  the  section    of  the   column    being   (a)    a  circle ;    (^)    a  square. 
£  =  29,000,000  Ibs.,  and  the  stress  due  to  direct  thrust  =  1500  Ibs.  per 
square  inch.  Ans. — (a)  30,615  Ibs.;  (£)  26,715  Ibs. 

40.  With  the  same  maximum    stress  as  in    the  last   question,  find 
the  angular  deviation  at  the  ends  so  that  the  stress  due  to  direct  thrust 
may  be  10,000  Ibs.  per  square  inch.  Ans. — (a)  i°  5';  (b)  i°  33'. 

41.  Show  that  the  load  required  to  produce  an  angular  deviation  of 
14°  at  the  two  pin  ends  of  a  long  column  is  only  one  per  cent  greater 
than  that  which  just  produces  flexure. 


CHAPTER  IX. 
TORSION. 

I.  TORSION  is  the  force  with  which  a  thread,  wire,  or  pris- 
matic bar  tends  to  recover  its  original  state  after  having  been 
twisted,  and  is  produced  when  the  external  forces  which  act 
upon  the  bar  are  reducible  to  two  equal  and  opposite  couples 
(the  ends  of  the  bar  being  free),  or  to  a  single  couple  (one  end 
of  the  bar  being  fixed),  in  planes  perpendicular  to  the  axis  of 
the  bar.  The  effect  upon  the  bar  is  to  make  any  transverse 
section  turn  through  an  angle  in  its  own  plane,  and  to  cause 
originally  straight  fibres,  as  DE,  to  assume  helicoidal  forms,  as 
FG  or  DC.  This  induces  longitudinal  stresses  in  the  fibres, 


/p     p\ 


p  p  'f 

I  0 

FIG.  360.  FIG.  361. 

and  transverse  sections  become  warped.  It  is  found  suf- 
ficiently accurate,  however,  in  the  case  of  cylindrical  and  regu- 
lar polygonal  prisms,  to  assume  that  a  transverse  section  which 
is  plane  before  twisting  remains  plane  while  being  twisted. 
In  order  that  the  bar  may  not  be  bent,  its  axis  must  coincide 
with  the  axis  of  the  twisting  couple. 

2.  Coulomb's  Laws.— The  angle  turned  through  by  one 
transverse  section  relatively  to  another  at  a  unit  distance  from 
it,  is  called  the  Antrle  of  Torsion,  and  Coulomb  deduced  from 

568 


TOR  SIGNAL    STRENGTH   OF  SHAFTS.  569 

experiments  upon  wires,  that  this  angle  is  directly  proportional 
to  the  moment  of  the  twisting  couple,  and  inversely  propor^ 
tional  to  the  fourth  power  of  the  diameter. 

Thus,  if  a  force  P,  at  the  end  of  a  lever  of  radius  p,  twists  a 
cylindrical  bar  of  length  L  and  radius  R,  and  if  6  is  the  circu- 
lar measure  of  the  angle  of  torsion,  then 

0  oc  Pp,  and  also  oc  — -, 

so  that  6  =  C  -~^  C  being  a  constant  depending  only  upon  the 

nature  of  the  material. 

Let  T  be  the  total  angle  of  torsion,  in  circular  measure, 
i.e.,  the  angle  turned  through  by  one  end  of  the  bar  relatively 
to  the  other.  Then 

T  —  0  _  rPP 

3.  Torsional  Strength  of  Shafts  (see  Art.  23,  Chap. 
IV). — Consider  a  portion  of  the  shaft  bounded 
by  the  planes  CE  and  MN9  Fig.  361.  It  is  kept 
in  equilibrium  by  the  couple  (P,  —P\  and  by 
the  elastic  resistance  at  the  section  MN.  Hence, 
this  elastic  resistance  must  be  equivalent  to  a 
couple  equal  and  opposite  to  (P,  — P).  FIG.  362. 

Let  Fig.  362  be  the  transverse  section  at  MN9  on  an  en- 
enlarged  scale,  and  let  abb' a'  be  any  elementary  area  (=  4A^ 
(P9—P)  of  the  surface  bounded  by  the  radii  OA,  OB,  and  by 
the  concentric  arcs  aa' ,  bb' . 

Let  xl  be  the  distance  of  AAl  from  O. 

It  is  assumed,  and  is  approximately  true,  that  the  resist- 
ance of  any  element  abb' a'  to  torsion  is  directly  proportional 
to  the  angle  of  torsion  (#),  to  its  distance  from  the  axis  (#,), 
and  to  its  area  (^M,),  and  also  that  it  acts  at  right  angles  to 
the  radial  line  of  the  element,  i.e.,  to  OA  or  OB. 

Thus,  the  resistance  of  abb' a'  to  torsion  =  G9x^AA\^  G  be- 
ing a  constant  to  be  determined  by  experiment. 

The  corresponding  moment  of  resistance  about  the  axis  = 
14,.       Similarly,    if    x^  jr8,  x^9    ...    are    the    distances 


5/0  THEORY   OF  STRUCTURES. 

from  the  axis  of  any  other  elements,  4AZ,  4Aa,  4A4,  .  .  .  t 
respectively,  the  corresponding  moments  of  resistance  are 
Gtix*AA^>  G8x*dAz,  .  .  .  Hence,  the  total  moment  of  resist- 
ance of  the  section 


=  G6I, 

I  being  the  moment  of  inertia  with  respect  to  the  axis. 
But  this  moment  of  resistance  (M)  is  equal  and  opposite  to 
the  moment  of  the  couple  (P,—P).     Hence, 

M=GOI=Pp. 

The  twisting  moment  will  of  course  vary  with  a  variable 
resistance,  and  the  last  equation  gives  its  mean  value. 

The  shaft,  however,  must  be  designed  (see  Cor.  4)  for  the 
maximum  couple  to  whi<*h  it  may  be  subjected,  and  the  moment 
of  this  couple  (=  J/,)  may  be  expressed  in  terms  of  the  mean 
by  the  equation 


ju.  being  a  coefficient  to  be  determined  in  each  case.  In  a  series 
of  experiments  with  different  engines,  Milton  found  that  // 
varied  from  1.3  to  2.1,  but  doubtless  the  variation  is  often  be- 
tween still  wider  limits. 

Cor.  i.  Let  /be  the  stress  at  the  point  farthest  from  the 
axis.     For  a  solid  round  shaft,  of  diameter  D, 

and    =G9». 


32 


Let  T°  be  the  total  torsion  in  degrees.     Then 

i  nT°    , 
z 


J^OR  SIGNAL   STRENGTH   OF   SHAFTS. 

and  hence 

D 


"  L   1  80   2' 
or 


_ 
D~  fi.60 

Taking  the  following  mean  values  of  G 

Material.  G  f 

Cast-iron  ..............   6,300,000  5,6oo 

Wrought-iron  ..........  10,500,000  7,200 

Steel  ..................  12,000,000  1  1,200 

2j  =  9.8T°  for  cast  iron,  =  12.7^°  for  wrought-iron,  =  9.3^° 

for  steel. 

Thus,  the  twist  is  i°  each  9.8  diameters  in  length  for  cast- 
iron,  each  12.7  diameters  in  length  for  wrought-iron,  and  each 
9.3  diameters  in  length  for  steel.  This  is  often  much  too  small, 
and  in  practice  the  twist  is  usually  limited  to  T^-°  per  lineal  foot 
of  length.  For  a  hollow  round  shaft,  D  being  the  external  and 
Dl  the  internal  diameter, 


If  the  thickness  (7")  of  the  hollow  shaft  is  small  compared 
with  D, 

&  -  Z>34  =  D'  -  (D  -  2Ty  =  SD*T,  approximately, 

•  \ 
and 


5/2  THEORY  OF  STRUCTURES. 

The  use  of  compressed  steel  admits  of  shafts  being  made 
hollow.     For  a  solid  square  shaft,  H  being  the  side  of  the  square, 


IT 

and/,  the  stress  at  the  end  of  a  diagonal,  =  G6—-=- 


H     6 


and 


GI      GH* 


In  these  results  it  is  assumed  that  G0[  —  —  or  =  — =r~  I 

\       D  H  / 

is  constant  at  different  points  of  the  cross-section,  which,  how- 
ever, is  only  true  for  circular  sections. 

In  non-circular  sections  the  stress  is  more  generally  greatest 
at  points  in  the  bounding  surface  which  are  nearest  to  the  axis 
and  least  at  those  points  which  are  farthest  from  the  axis. 
St.  Venant,  who  first  called  attention  to  this  fact,  gave  the  fol- 
lowing, amongst  others,  as  the  results  of  his  investigations. 

Designating  by  unity  the  torsional  rigidity  \^=  —j-j  of  a  shaft 
with  circular  section,  the  torsional  rigidity  of  a  shaft  of  equal 

/      271  /~fr 

sectional  area  is  .8863,  .8863  X  \  /  n*  .  l ,  .7255,  or  \  /  -,  ac- 
cording as  the  section  is  a  square,  a  rectangle  with  sides  in 
the  ratio  n  to  I,  an  equilateral  triangle,  or  an  ellipse  whose 
major  and  minor  axes  are  2a  and  2b,  respectively. 

Cor.  2.  The  torsional  stress  per  unit  of  area-  at  a  distance  x 
from  the  axis  is  GOx. 

Hence,  if  9  —  i  and  x  =  i,  G  is  the  force  that  will  twist  a 


TORSIONAL    STRENGTH  OF  SHAFTS.  573 

unit  of  area  at  a  unit  of  distance  from  the  axis  through  an  angle 
unity. 

Cauchy  found  analytically  that  in  an  isotropic  body  G  is 
two-fifths  of  the  coefficient  of  direct  elasticity. 

Experiments  indicate  that  G  is  about  three-eighths  or  one- 
third  of  the  coefficient  of  direct  elasticity. 

f  -f  (1  r>4 

Cor.  3.   For  a  solid  cylinder,  Pp  =  -  ,  R  being  the  radius, 

and  therefore  R*  oc  -^T>     If  the  shaft  is  to  have  a  certain  speci- 
Gv 

fied  stiffness,  i.e.,  if  0  is  fixed,  R*  oc  -?r,  and  for  a  given  twisting 

moment  R*  oc  —  .     Now  G  is  nearly  the  same  for  wrought-iron 
G 

and  steel,  so  that  there  is  little  if  any  advantage  to  be  gained 
by  the  use  of  the  latter. 

After  passing  the  elastic  limit,  the  stress  varies  much  more 
slowly  than  as  the  distance  from  the  axis,  and  there  will  be  a 
partial  equalization  of  stress,  the  apparent  torsional  strength 
being  increased. 

Cor.  4.  In  any  transverse  section  of  a  solid  cylindrical  shaft, 
the  maximum  unit  stress 


Ml  being  the  moment  of  the  maximum  twisting  couple. 

This  relation  is  true  so  long  as  the  stress  does  not  exceed 
the  elastic  limit,  and  agrees  with  the  practical  rule  that  the 
diameter  of  a  cylindrical  shaft  subjected  to  torsional  forces  is 
proportional  to  the  cube  root  of  the  twisting  couple. 

The  rule  is  usually  expressed  in  the  form 

M,  =  KD\     so  that     K  =  ^. 

Wohler's  experiments  show  that  the  value  of  f  depends, 
to  some  extent,  upon  its  fluctuation  under  the  variable  twist- 


574  THEORY   OF  STRUCTURES. 

ing  moment.  Ordinarily  it  should  not  exceed  7200  Ibs.  per  square 
inch  for  wrought-iron,  in  which  case  K  —  i\  J^  X  --f-  =  I4H- 
(Note.  —  If  Pl  is  the  torsional  breaking  weight, 


_ 

"  D*  ~~  I}3 


is  the  coefficient  of  torsional  rupture?) 

Cor.  5.  Let*  Wbe  the  work  transmitted  to  a  shaft  of  D  in. 
diameter,  in  foot-pounds  per  minute,  N  being  the  correspond 
in«-  number  of  revolutions.  Then 

M 

12  W  =  inch-pounds  transmitted  =  2nMN  —  2it  —TV 


KD 


Ml 
since  M  =  mean  twisting  moment  =  —  -.     Hence, 


w    « 


Let  HP  be  the  horse-power  transmitted  per  minute.    Then 
W  =  33000  HP.     Also  for  wrought-iron  K  —  i|££  x  -?f. 


Hence  p—  --  -^-  =  D\     and  if     p  =  1.43, 


HP 


N 


-i  formula  agreeing  with  the  best  practice  in  the  case  of 
wrought-iron  shafts  subjected  to  torsional  forces  only.  Such 
shafts  should,  therefore,  carry  no  pulleys. 

Cor.  6.  The  resilience  of  a  cylindrical  axle  is  the  product  of 
one  half  of  the  greatest  moment  of  torsion  into  the  correspond- 
ing angle  of  torsion. 

Cor.  7.  It  often  happens  in  practice  that  a  shaft  (or  beam) 
is  subjected  to  a  bending  as  well  as  to  a  torsional  action. 


DISTANCE   BETWEEN    THE  BEARINGS   OF   SHAFTING.   $?$ 

The  combined  bending  and  twisting  moments  are  equiv- 
alent (Art.  8,  Chap.  IV)  to  the  moment 


where  Mb  =  nMt,  Mb  being  the  bending  and  Mt  the  twisting 
moment  at  the  given  section. 

Hence,  remembering  that  the  maximum  twisting  moment 
J/,  is  equal  to  pMtJ  we  have  for  a  wrought-iron  shaft, 


If  n  =  .36  -f-  ,  this  becomes 


HP 


a  formula  agreeing  with  the  best  practice  in  the  case  of  trans- 
mission with  bending,  as,  e.g.,  in  the  crank-shafts  of  marine 
engines. 

It  often  happens  that  n  has  a  still  larger  value,  as,  e.g.,  in 
the  case  of  head  shafts  properly  supported  against  springing. 
The  usual  formula  is  then 


D 


3     I  HP 

=  5V  W' 


corresponding  to  n  =  .72  -{- . 

4.  Distance  between  Bearings. — The  distance  between 
the  bearings  of  a  line  of  shafting  is  limited  by  the  considera- 
tion that  the  stiffness  of  the  shaft  must  be  such  as  will  enable 
it  to  resist  excessive  bending  under  its  own  weight  and  under 
any  other  loads  (e.g.,  pulleys,  wheels,  etc.)  applied  to  it. 
For  this  reason,  the  ratio  of  the  maximum  deviation  of  the 
axis  of  the  shaft  from  the  straight  to  the  corresponding  dis- 
tance between  bearings  should  not  exceed  a  certain  fraction 
whose  value  has  been  variously  estimated  by  different  writers. 

Let  /  be  the  distance  in  feet  between  bearings,  a  the 
diameter  of  the  shaft  in  inches,  w  the  weight  of  the  ma- 


576  THEORY  OF  STRUCTURES. 

terial  of  the  shaft  per  cubic  foot,  and  let  the  applied  load  be 
equivalent  to  a  load  per  lineal  unit  of  length  m  times  that 
of  the  shaft.  Assume  a  stiffness  of  T^Vir»  an^  tnat  the  axis  of 
the  shaft  is  truly  in  line  at  the  bearings.  The  maximum  de- 
flection of  the  shaft  is  given  by  the  formula  (Art.  3,  Ex.  8. 
Chap.  VII) 

_      I    (m  +  i)(weight  of  shaft)/3  .  1728 

-  ~ 


I  nd*    i          64     /3  .  1728 

-s-UH  +  l)  --  tt>/-~r  -  ~  -  . 

384  v  4    144        ar*/4        ^ 


/  "     IOO  2E 

or 


_   s  /       Ed* 

~  •  V   5ow(m  +  0 


EXAMPLE. — For  wrought-iron,    E  =  3o,ocxD,ooo  Ibs.  and 
w  —  480  Ibs. 


If  the  applied  load,  instead  of  being  uniformly  distributed 
is  concentrated  at  the  centre,  the  maximum  deflection 

i    (m  +  j)( weight  of  shaft)/3 .  1728 
~  192  El  ~~' 

and  hence 


~,          Ed' 


[QOw(m  -f-  i 
EXAMPLE. — For  wrought-iron  /=  8.5 


CYLINDRICAL   SPIRAL    SPRINGS. 


577 


5.  Efficiency  of  Shafting.  —  Let  it  require  the  whole  of 
the  driving  moment  to  overcome  the  friction  in  the  case  of  a 
shaft  of  diameter  d  and  length  L.  The  efficiency  of  a  shaft  of 

the  same  diameter  and  length  /  =  I  —  -=•  . 


But 


,.     =  (Pp)  =  moment  of  friction  =  /*—  —  L— 


w  being  the  specific  weight  of  the  material  of  the  shaft,  and 
the  coefficient  of  friction.     Hence, 


w 


wl 
and  the  efficiency  =  I  —  2yw--. 


Hence, 


S=  2nyn,  approx. 


Wy  = 


fnr*       GQnr* 


2  2 

r  being  the  radius  of  the  spring. 


6.   Cylindrical   Spiral  Spring. — Let  the  figure  represent 
a  cylindrical  spiral  spring  of  length  s,  supporting 
a  weight  W.      Consider  a  section  of  the  spring  at 
any  point  B. 

At  this  point  there  is  a  shear  W^and  a  torque 
Wy,  y  being  the  distance  of  B  from  the  axis  of 
the  spring,  i.e.,  the  radius  of  the  coil. 

The  effect  of  W  may  generally  be  neglected 
as  compared  with  the  effect  of  the  moment  Wy, 
and  it  may  be  therefore  assumed  that  the  spring 
is  under  torsion  at  every  point.  Let  there  be  n 
coils.  Then 


FIG.  363. 


578  THEORY   OF   STRUCTURES. 

The  elongation  of  the  spring 

Syf      2nfny* 


__ 

:  ~~ 


Wy  a          Wy'S       fnr'S 
The  work  done  —OS  =    - 


A  weight  hung  at  the  lower  end  tends  to  turn  as  well  as 
lengthen  the  spring,  and  this  is  due  to  a  slight  bending  action. 

According  to  Hartnell,  f  •=.  60,000  Ibs.  per  square  inch  for 
•f-in.  steel,  f  =  50,000  Ibs.  per  square  inch  for  -J-in.  steel,  and 
G  varies  from  13,000,000  Ibs.  for  J-in.  steel  to  11,000,000  Ibs. 
for  f-in.  steel. 

Also  for  wire  less  than  f  in.  in  diameter, 

Wny* 
and  the  deflection  — 


EXAMPLE. — A  wrought-iron  shaft  in  a  rolling-mHl  makes  50 
revolutions  per  minute  and  transmits  120  H.  P.,  which  is  sup- 
plied from  a  waterfall  by  means  of  a  turbine.  Determine  the 
diameter  of  the  shaft  (i)  if  the  maximum  stress  in  the  metal 
is  not  to  exceed  9000  Ibs.  per  square  inch ;  (2)  if  the  angle  of 
torsion  is  not  to  exceed  ^°  per  lineal  foot. 

As  a  matter  of  fact,  the  diameter  of  the  shaft  is  3 j  in.  at 
the  bearings  and  4  in.  in  the  intermediate  lengths.  What  are 
the  corresponding  maximum  inch-stresses  in  the  metal? 

Let  the  twisting  couple  be  represented  by  force  P  at  the 
end  of  an  arm  /.  Then 

P  X  27tp  x  95  =  120  X  33,ooo  ft.-lbs. 

120  X  33000    126000  ,          126000 

.%  Pp  =  ll4£-_  _  —  _      _  ft.-lbs.  = — X  12  in.-lbs. 

27r  X  95  19  J9 

126000  X  12 
First,  -    —  - 

and  hence 


EXAMPLE.  579 


,  126000 
Second,  --  X  12  =  Pp  = 


But  6  --  ~  X  y-  X  ~  ;    take  G  =  io,5OO,cxx).     Then 

126000  X    12         10500000         22  I  I  I  22  „ 

~  T9  --  =  —  2—  x  7X  785  x  n  X77  x  7^- 

Hence, 

Z^4  =689.45,     and    D=  5.12  in. 

Third,  the  maximum  stresses  in  the  real  shaft  at  the  bear- 
ings and  in  the  intermediate  lengths  are  respectively  given  by 

126000  stress      22 

_XI2  =  __X_ 

and 

126000  stress       22 

—^-  X  12,    -^-  Xy  X(4). 

From  the  former,  the  maximum  stress  =  7682  Ibs.  persq.  inch. 
"    latter,      "  "  «      =6330    "      "     "      " 


580  THEORY  OF  STRUCTURES. 


EXAMPLES. 

1.  A  steel  shaft  4  in.  in  diameter  is  subjected  to  a  twisting  couple 
which  produces  a  circumferential  stress  of  15,000  Ibs.    What  is  the  stress 
(shear)  at  a  point  I  in.  from  the  centre  of  the  shaft  ? 

Determine  the  twisting  couple.  Ans.  7500  Ibs.;  23,57 if  Ibs. 

2.  A  weight  of  2|  tons  at  the  end  of  a   i-ft.  lever  twists  asunder  a 
steel  shaft  if  in.  in  diameter.     Find  the  breaking  weight  at  the  end  of  a 
2-ft.  lever,  and  also  the  modulus  of  rupture. 

Ans.   1 1  tons  ;  23, 510  Ibs. 

3.  A  couple  of  A^ft.-tons  twists  asunder  a  shaft  of  diameter  d.    Find 
the  couple  which  will  twist  asunder  a  shaft  of  the  same  material  and 
diameter  2.d.  Ans.  &N. 

4.  Compare  the  couples  required  to  twist  two  shafts  of   the  same 
material  through   the  same  angle,  the  one  shaft  being  /  ft.  long  and 
d  in.  in  diameter,  the  other  2/  ft.  long  and  id  in.  in  diameter. 

Compare  the  couples,  the  diameter  of  the  latter  shaft  being  — . 

Ans.   i  to  8 ;  32  to  I. 

5.  A  shaft  1 5  ft.  long  and  4^  in.  in  diameter  is  twisted  through  an  angle 
of  2°  under  a  couple  of  2000  ft.-lbs.     Find  the  couple  which  will  twist 
a  shaft  of  the  same  material  20  ft.  long  and  i\  in.  in  diameter  through 
an  angle  of  2^°.  Ans.   12,288  ft.  Ibs. 

6.  A  round  cast-iron  shaft  15  ft.  in  length  is  acted  upon  by  a  weight 
of  2000  Ibs.  applied  at  the  circumference  of  a  wheel  on  the  shaft ;  the 
diameter  of  the  wheel  is  2  ft.     Find  the  diameter  of  the  shaft  so  that  the 
total  angle  of  torsion  may  not  exceed  2°.  Ans.  3.53  in. 

7.  A  wrought-iron  shaft  is  subjected  to  a  twisting  couple  of  12,000  ft.- 
lbs.  ;  the  length  of  the  shaft  between  the  sections  at  which  the  power  is 
received  and  given  off  is  30  ft.;  the  total  admissible  twist  is  4°.     Find 
the  diameter  of  the  shaft,  ju  (page  570)  being  f ,  and  m  10,000,000  Ibs. 

Ans.  7.74.  in. 

8.  A  wrought-iron  shaft  20  ft.  long  and    5  in.  in  diameter  is  twisted 
through  an  angle  of  2°.     Find  ihe  maximum  stress  in  the  material,  m 
being  10,500,000  ft.-lbs.  Ans.  3819.2  ibs.  per  sq.  in. 


EXAMPLES.  58l 

9.  A  crane  chain  exerts  a  pull  of  6000  Ibs.  tangentially  to  the  drum 
upon  which  it  is  wrapped.      Find  the  diameter  of  a  wrought-iron  axle 
which  will  transmit  the  resulting  couple,  the  effective  radius  of  the  drum 
being  7^  in. ;  the  safe  working  stress  per  square  inch  being  7200  Ibs. 

Ans.  3.17  in. 

10.  Find  the  diameter  and  the  total  angle  of   torsion  of  a   12-ft. 
wrought-iron  shaft  driven   by  a  water-wheel  of  20  H.  P.,  making   25 
revolutions  per  minute,  m  being  10,000,000  Ibs.,  and  the  working  stress 
7200  Ibs.  per  square  inch.  Ans.   5.6  in.;  2°. 2. 

n.  A  turbine  makes  1 14  revolutions  per  minute,  and  transmits  92 
H.  P.  through  the  medium  of  a  shaft  8  ft.  6  in.  in  length.  What  must  be 
the  diameter  of  the  shaft  so  that  the  total  angle  of  torsion  may  not  ex- 

2° 

ceed  — ,  m  being  10,500,000  Ibs.  ?  Ans.  4.7  in. 

Determine  the  side  of  a  square  .pine  shaft  that  might  be  substituted 
for  the  iron  shaft. 

12.  A  steel  shaft  20  ft.  in   length   and  3  in.  in  diameter  makes  200 
revolutions  per  minute  and  transmits  50  H.  P.     Through  what  angle  is 
the  -shaft  twisted  ? 

A  wrought-iron  shaft  of  the  same  length  is  to  do  the  same  work  at 
the  same  speed.  Find  its  diameter  so  that  the  stress  at  the  circumference 
may  not  exceed  f  of  that  at  the  circumference  of  the  steel  shaft. 

Ans.   2°. 6;  3.556  in. 

13.  A  vertical  cast-iron  axle  in  the  Saltaire  works  makes  92  revolu- 
tions per  minute  and  transmits  300  H.  P.;  its  diameter  is  10  in.     Find 
the  angle  of  torsion.  Ans.  .0144°  per  lineal  foot. 

14.  In  a  spinning-mill  a  cast-iron  shaft  8J  in.  in  diameter  makes  27 
revolutions  per  minute;  the  angle  of  torsion  is  not  to  exceed  — per 
lineal  foot.     Find  the  work  transmitted.  Ans.  62.19  H.  P. 

15.  A  square  wooden  shaft  8  ft.  in  length  is  acted  upon  by  a  force  of 
200  Ibs.,  applied  at  the  circumference  of  an  8  ft. -wheel  on  the  shaft. 
Find  the  length  of  the  side  of  the  shaft,  so  that  the  total  torsion  may 
not  exceed  2°  (in  =  400000).     What  should  be  the  diameter  of  a  round 
shaft  of  equal  strength  and  of  the  same  material  ? 

Ans.  4.96  in.;  5.09  in. 

16.  A  shaft  transmits  a  given  H.  P.  at  ^revolutions  per  minute  with- 
out bending.     Find  the  weight  of  the  shaft  in  pounds'per  lineal  foot. 

/H.P.M 

Ans.     2.-     . 


582  THEORY  OF  STRUCTURES. 

17.  The  working  stress  in  a  steel  shaft  subjected  to  a  twisting  couple 
of  looo  in.-tons  is  limited  to  11,200  Ibs.  per  square  inch.     Find  its  diam- 
eter; also  find  the  diameter  of  the  steel  shaft  which  will  transmit  5000 
H.  P.  at  66  revolutions  per  minute,//  being  f.         Ans.  10  in.  ;  6.88  in. 

1 8.  A  wrought-iron  shaft  is  twisted  by  a  couple  of  10  ft. -tons.     Find 
its  diameter  (a)  if  the  torsion  is  not  to  exceed  i°  per  lineal  foot,  (ft)  if  the 
safe  working  stress  is  7200  Ibs.  per  square  inch,     m  =  10,000,000  Ibs. 

Ans.—  (a)  3.7  in.;  (&)  5.7  in. 

19.  A  steel  shaft  2  in.  in  diameter  makes  100  revolutions  per  minute 
and  transmits  25  H.  P.     Find  the  maximum  working  stress  and  the  tor- 
sion per  lineal  foot,  m  being  10,000,000  Ibs.     Also  find  the  diameter  of  a 
shaft  of  the  same  material  which  will  transmit  100  H.  P.  with  the  same 
maximum  working  stress.  Ans.  io,o22T\  Ibs. ;  .0478°  ;  3.17  in. 

20.  The  crank  of  a  horizontal  engine  is  3  ft.  6  in.  and  the  connecting- 
rod  9  ft.  long.     At  half-stroke  the  pressure  in  the  connecting-rod  is  500 
Ibs.     What  is  the  corresponding  twisting  moment  on  the  crank-shaft  ? 

Ans.  1716^  ft.-lbs. 

21.  If  the  horizontal  pressure  upon  the  piston  end  of  the  connecting 
rod  in  the  previous  question  is  constant,  find  the  maximum  twisting  mo- 
ment on  the  crank-shaft. 

/  .  sin  0cos0    \ 

Ans.  Pi  sin  0  +  — _         --  j ,    0  being  given  by 

n*  cos20  +  «4(sin2  0  cos2  0  —  i)  +  «2  sin4  0(i  +  sin2  0)  —  sin  <  0  =  o 

where  n  =  *£/-  =  \8-. 
N.B. — If  sin20  is  neglected  as  compared  with  «2, 

the  maximum  moment  =  P  sin  0(1+  — 

V  n 

0  being  very  nearly  72°. 

22.  Show  that  a  hollow  shaft  is  both  stiffer  and  stronger  than  a  solid 
shaft  of  the  same  weight  and  length. 

23.  Find  the  percentage  of  weight  saved  by  using  a  hollow  instead  of 
a  solid  shaft. 

Ans.  If  of  equal  stiffness  =  — . 

m*  +  i 

If  of  equal  strength  =  100  \  i  —  /Jm\m*  ~  0   ! 
1  («"  +  Oa     ) 

m  being  the  ratio  of  the  external  to  the  internal  diameter 
of  hollow  shaft. 

24.  A  hollow  cast-iron  shaft  of  12  in.  external  diameter  is  twisted  by 
a  couple  of  27,000  ft.-lbs.     Find  the  proper  thickness  of  the  metal  so  that 
the  stress  may  not  exceed  5000  Ibs.  per  square  inch.  Ans.  .619  in. 


EXAMPLES. 

25.  The  external  diameter  of  a  hollow  shaft  is/  times  the  internal. 
Compare  its  torsional  strength  with  that  of  a  solid  shaft  of  the  same  ma- 
terial and  weight. 


Ans.  -— 

/J  +  i 

26.  If  the  solid  shaft  is  10  in.  in  diameter,  and  the  internal  diameter 
of  the  hollow  shaft  is  5  inches,  find  the  external  diameter  and  compare 
the  torsional  strengths.  AnSt  5  j/g  in<  .  ^  to  3. 

27.  A  hollow  steel  shaft  has  an  external  diameter  d  and  an  internal 

d 
diameter  —  .     Compare  its  torsional  strength  with  that  of  (a)   a  solid 

steel  shaft  of  diameter  d\  (b)  a  solid  wrought-iron  shaft  of  diameter  d\ 
the  safe  working  stresses  of  steel  and  iron  being  5  tons  and  3!  tons  re- 
spectively. Ans—  (a)  -fa  ;  (£)  |-|. 

28.  What  twisting  moment  can  be  transmitted  by  a  hollow  steel  shaft 
of  8  in.  internal  and  10  in.  external  diameter,  the  working  stress  being 
$  tons  per  square  inch  ?  Ans.  184^  in.-tons. 

29.  If/i  is  the  safe  torsional  working  stress  of  a  shaft,  and  /a  is  the 
safe  working  stress  when  the  shaft  acts  as  a  beam,  show  that  the  tor- 
sional resistance  of  the  shaft  is  to  its  bending  resistance  in  the  ratio  of 

2/a  tO/,. 

30.  The  wrought-iron  screw  shaft  of  a  steamship  is  driven  by  a  pair 
of  cranks  set  at  right  angles  and  21.7  in.  in  length;  the  horizontal  pull 
upon  each  crank-pin  is  176,400  IDS.,  and  the  effective  length  of  the  shaft 
is  866  in.     Find  the  diameter  of  the  shaft  so  that  (i)  the  circumferential 
stress  may  not  exceed  9000  Ibs.  per  square  inch  ;  (2)  the  angle  of  torsion 

rd 
may  not  exceed  —  per  lineal  foot  ;  m  being  10,000,000  Ibs.     The  actual 

diameter  of  the  shaft  is  14.9  in.     What  is  the  actual  torsion  ? 

Ans.  —  (i)  14.53  m'J  (2)  H-89  in.;  (3)  total  torsion  =  5°.  545. 

31.  The  ultimate  tensile  strength  of  the  iron  being  60,000  Ibs.  per 
square  inch,  find  the  actual  ultimate  strength  under  unlimited  repetitions 
of  stress.  Ans.  54,899  Ibs.  (Unwin's  formula). 

32.  What  is  the  torsion  in  the  preceding  question  when  one  of  the 
cranks  passes  a  dead  point  ? 

33.  A  steel  shaft  300  feet  in  length  makes  200  revolutions  per  minute 
and  transmits  10  H.  P.    Determine  its  diameter  so  that  the  greatest 
stress  in  the  material  may  be  the  same  as  the  stress  at  the  circumference 
of  an  iron  shaft  i  in.  in  diameter  and  transmitting  500  ft.  -Ibs. 

Ans.  .807  in.  (=  £  in.) 

34.  Determine  the  coefficient  of  torsional  rupture  for  the  shaft  in 
Question  33,  10  being  the  factor  of  safety. 

35.  A  wrought-iron  shaft  in  a  rolling-mill  is  220  feet  in  length,  makes 
95  revolutions  per  minute,  and  transmits  1  20  H.  P.  to  the  rolls;  the  main 
body  of  the  shaft  is  4  in.  in  diameter,  and  it  revolves  in  gudgeons  3f  in. 


584  THEORY  OF  STRUCTURES. 

in  diameter.     Find  the  greatest  shear  stress  in  the  shaft  proper  and  in 
the  portion  of  the  shaft  at  the  gudgeons.       Ans.  6330.2  Ibs.;  7508  Ibs. 

36.  Power  is  taken  from  a  shaft  by  means  of  a  pulley  24  inches  in 
diameter  which  is  keyed  on  to  the  shaft  at  a  point  dividing  the  distance 
between  two  consecutive   supports  into  segments  of  20  and  80  in.  ;  the 
tangential  force  at  the  circumference  of  the  pulley  is  5500  Ibs.     If  the 
shaft  is  of  cast-iron,  determine  its  diameter,  taking  into  account  the 
bending  action  to  which  it  is  subjected.  Ans.  4.7  in. 

37.  Show  that  the  resilience  of  a  twisted  shaft  is  proportional  to  its 

weight.  /2  Volume 

Ans.   Resilience  = . 

in         4 

38.  If  a  round  bar  of  any  material  is  subjected  to  a  twisting  couple, 
show  that  its  maximum  resilience  is  two-thirds  the  maximum  resilience 
of  the  material. 

39.  Determine  the  diameter  of  a  wrought-iron   shaft   for  a  screw 
steamer,  and  the  torsion  per  lineal  foot;  the  indicated  H.  P.  =  1000,  the 
number  of  revolutions  per  minute  =  150,  the  length  of  the  shaft  from 
thrust  bearing  to  screw  =  75  ft.,  and  the  safe  working  stress  =  7200  Ibs. 
per  square  inch.  Ans.  6.67  in.  ;  io°.5. 

40.  In  a  spinning-mill  a  cast-iron  shaft  84  ft.  long  makes  50  revolu- 
tions per  minute  and  transmits  270  H.  P.     Find  its  diameter  (i)  if  the 
stress  in  the  metal  is  not  to  exceed  5000  Ibs.  per  square  inch ;  (2)  if  the 

,        o 

angle  of  torsion  per  lineal  foot  is  not  to  exceed  — . 

Also  (3)  in  the  first  case  find  the  total  torsion. 

Ans.   (i)  7.02  in. ;  (2)  10.23  in.  ;  (3)  28°. 8. 

41.  A  circular  shaft  is  twisted  beyond  the  limit  of  elasticity.     If  the 
equalization  of  stress  is  perfect,  show  that  for  a  given  maximum  stress  the 
twisting  couple  is  greater  than  it  would  be  if  the  elasticity  were  perfect, 
in  the  ratio  of  4  to  3. 

42.  Determine  (a)  the  profile  of  a  shaft  of  length  /  which  at  every 
point  is  so  proportioned  as  to  be  just  able  to  bear  the  power  it  has  to 
transmit  plus  the  power  required  to  overcome  the  friction  beyond  the 
point  under  consideration.     Find  (ft)  the  efficiency  of  such  a  shaft,  and 
(c)  the  efficiency  of  a  shaft  made  up  of  a  series  of  n  divisions,  each  of 
uniform  diameter. 

Ans.  (a)  The  radius/  of  any  section  distant  x  from  the  driving  end 

X 

is y  =  re  ?>L  ,  r  being  the  radius  of  the  driving  end  and 
L  the  length  of  a  shaft  of  uniform  diameter,  such  that 
the  whole  driving  moment  is  required  to  overcome  its 
own  friction. 


EXAMPLES.  585 

43.  A  steel  shaft  carries  a  5~ft.  pulley  midway  between  the  supports 
and  makes  6  revolutions  per  minute,  the  tangential  force  on  the  pulley 
being  500  Ibs.    Taking  the  coefficient  of  working  strength  at  11,200  Ibs. 
per  square  inch,  find  the  diameter  of  the  shaft  and    the  proper  distance 
between  the  bearings. 

44.  A  steel  shaft  4  inches  in  diameter  and  weighing  490  Ibs.  per  cubic 
foot  makes   100  revolutions  per  minute.     If  the  working  stress  in  the 
metal  is  11,200  Ibs.  per  square  inch,  find  the  twisting  couple  and  the  dis- 
tance to  which  the  work  can  be  transmitted  ;    the  coefficient  of  friction 
being  .05,  and  the  efficiency  of  the  shaft  f. 

Ans.   140,800  in.  -Ibs.  ;  8228^  ft. 

45.  If  the  shaft  is  of  steel,  and  if  the  loss  due  to  friction  is  20  per  cent, 
find  the  distance  to  which  work  may  be  transmitted,  ju  being  .05. 

Ans.  6582!  ft. 

46.  A  wrought-iron  shaft  220  ft.  between  bearings  and  4  in.  in  diam- 
eter  can   safely  transmit   120  H.  P.  at  the  rate  of  95  revolutions  per 
minute.     What  is  the  efficiency  of  the  shaft?     (jit  =  fa.)       Ans.  .976. 

47.  The  efficiency  of  a  wrought-iron  shaft  is  £  ;  the  working  stress  in 
the  metal  is  7200  Ibs.  per  square  inch  ;  the  coefficient  of  friction  is  .125. 
How  far  can  the  work  be  transmitted  ?  Ans.  4320  ft. 

48.  A  spring  is  formed  of  steel  wire  ;  the  mean  diameter  of  the  coils 
is  i  inch  ;  the  working  stress  of  the  wire  is  50,000  Ibs.  per  square  inch  ; 
the  elongation  under  a  weight  of  19/3-  Ibs.  is  2  inches;  the  coefficient  of 
transverse  elasticity  is  12,000,000  Ibs.    Find  the  diameter  of  the  wire  and 
the  number  of  coils. 

49.  Find  the  weight  of  a  helical  spring  which  is  to  bear  a  safe  load  of  6 
tons  with  a  deflection  of  I  inch,  G  being  12,000,000  Ibs.,  and  f  60.000  Ibs. 

50.  Find  the  time  of  oscillation  of  a  spring,  the  normal  displacement 
under  a  given  load  being  A.  AJ 

Ans.  TI\  ~. 

5  1  .  Find  the  deflection  under  the  weight  W  of  a  conical  helical  spring 
(a)  of  circular  section  ;  (<£)  of  rectangular  section,  the  radii  of  the  extreme 
coils  being  ji  and  ja  ,  and  the  radial  distance  from  the  axis  to  a  point  of 
the  spring  at  an  angular  distance  </>  from  the  commencement  of  the  spiral 

r2  —  R         <p 

being  given  by  the  relation  ---  =  -  .     (n  =  number  of  coils.) 
yi  —  yi       27zv/ 

.n(y,  +  yd(y?+yS)W    f      ny*lV 
Ans.  (a)  -  ~~Grr~  V         Gr*~'      ^  =  °  ^  ~ 


b  and  h  being  the  sides  of  the  rectangular  section. 
52.  The  efficiency  of  an  axle  is  \  ;  the  working  stress  in  the  shaft  is 
9000  Ibs.  per  square  inch  ;  the  coefficient  of  friction  is  .10.    How  far  may 
work  be  transmitted  ? 


585* 


THEORY  OF  STRUCTURES. 


B 


Fig.  A  shows  the  distortion  produced  by  twisting1  a  round  %-in.  iron  bar. 
Fig.  B  shows  the  distortion  produced  by  twisting  a  square  %-in.  iron  bar. 


DISTORTION  OF  IRON  BARS  BY   TWISTING.          585^ 


The  above  figures  show  the  distortion  produced  by  twisting  a  1%  x  «fc£"  iron  bar. 


CHAPTER   X. 
STRENGTH   OF   CYLINDRICAL   AND   SPHERICAL   BOILERS. 

i.  Thin  Hollow  Cylinders ;  Boilers ;  Pipes. 

Let  r  be  the  radius  of  the  cylinder. 
Let  t  be  the  thickness  of  the  metal. 
Let/  be  the  fluid  pressure  upon  each  unit 
of  surface. 

Let/^  be  the  tensile  or  compressive  unit 
stress,  according  as  p  is  an  internal  or  exter- 
FIG.  364.  nal  pressure. 

Assume  (i)  that  the  metal  is  homogeneous  and  free  from 
initial  strain  ; 

(2)  that  /  is  small  as  compared  with  r ; 

(3)  that   the   pressures   are   uniformly    distributed 

over  the  internal  and  external  surfaces  ; 

(4)  that  the  ends  are  kept  perfectly  flat  and  rigid  ; 

(5)  that  the  stress  in  the  metal  is  uniformly  dis- 

tributed over  the  thickness. 

The  last  assumption  is  equivalent  to  supposing  that  it  is 
the  mean  circumferential  stress  which  is  governed  by  the 
strength  of  the  metal,  while  in  reality  it  is  the  internal  or  maxi- 
mum circumferential  stress  which  is  so  governed. 

The  figure  represents  a  cross-section  of  the  cylinder  of 
thickness  unity. 

A  section  made  by  any  diametral  plane,  as  AB,  must  de- 
velop a  total  resistance  of  2tf,  and  this  must  be  equal  and 
opposite  to  the  resultant  of  the  fluid  pressure  upon  each  half, 
i.e.,  to  2pr.  Hence, 

2tf=2pr,     or     tf=pr (i) 

586 


THIN  HOLLOW   CYLINDERS;    BOILERS;    PIPES.         587 

This  formula  may  be  employed  to  determine  the  bursting, 
proof,  or  working  pressure  in  a  cylindrical  or  approximately 
cylindrical  boiler,  provided  that/",  instead  of  being  the  tensile 
or  compressive  unit  stress,  is  some  suitable  coefficient  which 
has  been  determined  by  experiment.  If  r/  is  the  efficiency  of 
a  riveted  joint,  the  formula 


may  be  employed  to  determine  the  working  pressure  in  a  cylin- 
drical or  approximately  cylindrical  boiler. 

In  ordinary  practice  the  values  of  rf  and  /are  given  by  the 
following  table : 


Material. 

Joint. 

it 

/in  Ibs.  per  sq.  in. 

Wrought-iron  

Single-riveted 

-   *!*> 

8000  to  oooo 

Double-riveted 

•7 

H 

Treble-riveted 

8  to    85 

,, 

Steel  

Single-riveted 

.  ^^ 

I2OOO  to  I3OOO 

14 

Double-riveted 

7 

(, 

Treble-riveted 

8  to    85 

tt 

For  cast-iron  cylinders  the  working  value  of  /may  be  taken 
at  about  2000  Ibs.  per  square  inch. 

The  total  pressure  upon  each  of  the  flat  ends  of  the  cylinder 


The  longitudinal  tension  in  a  thin  hollow  cylinder 

K^P       P^     "    , 
"  2nrt  ~  2t'     ^  } 

and  is  one  half  of  the  circumferential  stress/. 

Cor.  i.  Let  the  cylinder  be  subjected  to  an  external  pressure 
p'  as  well  as  to  an  internal  pressure  p.     Then 


fl=pr-p'r>, 


(3) 


r'  being  the  radius  of  the  outside  surface  of  the  cylinder,    /is 
a  tension  or  a  pressure  according  as  pr 


588  THEORY  OF  STRUCTURES. 

Generally,  r  —  r'  is  very  small,  and  the  relation  (3)  may  be 
written 

ft  =  r(p-p'). 

2.  Thick  Hollow  Cylinder.  —  If  /  is  large,  the  stress  is  no 
longer  uniformly  distributed  over  the  thickness.  Suppose  that 
the  assumptions  (i)  and  (3)  of  Art.  I  still  hold,  also  that  the 
cylinder  ends  are  free,  and  that  the  annulus  forming  the  section 
of  the  cylinder  is  composed  of  an  infinite  number  of  concentric 
rings.  Under  these  conditions  the  straining  of  the  cylinder 
cannot  affect  its  cylindrical  form.  Hence,  right  sections  of  the 
cylinder  in  the  unstrained  state  remain  planes  after  the  strain- 
ing, so  that  the  longitudinal  strain  at  every  point  must  be  the 
same.  Two  methods  will  be  discussed. 

FIRST  METHOD.  —  Let  dx  be  the  thickness  of  one  of  the 
rings  of  radius  x,  and  let  dq  be  the  intensity  of  the  circum- 
ferential stress. 

pr  —  p'r'  =  difference   between   the   total    pressures   from 

within  and  without  =  total  circumferential  stress  =    /  *"  dq. 

If  it  be  assumed  that  the  thickness  (=  r'  —  r)  remains  un- 
changed under  the  pressure,  then  the  circumferential  extension 
of  each  of  the  concentric  rings  must  be  equal  to  the  same  con- 
stant quantity  A,  and  therefore 

dq  =  Edx  -  , 

* 


E  being  the  coefficient  of  elasticity.     Hence, 

.  ,       E\    Cr'dx      £1        r' 
pr—p'r'=-       I      —  =  —  log,-. 

27t  J  r       X  27t        *'  r 


en  /=  E 
2 

elastic  limit  is  not  exceeded,  and  therefore 


Let  /  be  the  tensile  unit  stress.     Then  /=  E  -  if  the 

2nr 


THICK  HOLLOW   CYLINDER.  589 

or 


r'  pr-p'r'       i(pr 

,.  -=  x  +  £_£-  +  -(—  -f-    ,  approx.,        (4) 

if  p  is  small  as  compared  with/;  and  hence, 

t        r'  pr-p'r'       ilr 

=     ~  ~~      h 


r 

In  most  cases  which  occur  in  practice/'  is  so  small  as  com- 
pared with/  that  it  may  be  disregarded. 
Hence,  making/'  zero  in  equation  (5), 


» 


Formulae  (5)  and  (6)  may  be  employed  even  if  the  elastic 
limit  is  exceeded,  if  f  is  considered  a  coefficient  of  strength 
to  be  determined  by  experience. 

Cor.  —  Rankine,  in  his  Applied  Mechanics,  obtains  by 
another  method, 


//±_P 

'-~\  f-P' 


if/'  be  neglected.     Hence, 


,  *3 

==  i  +y  +  2  7^  '  aPProximately' 

if  p  is  small  as  compared  with  /,  and  therefore 


59O  THEORY   OF  STRUCTURES. 


an  equation  identical  with  (6). 

SECOND  METHOD.  —  Consider  a  ring  bounded  by  the  radii 
x,  x  +  dx,  at  any  point. 

Let  q  be  the  normal  (i.e.,  radial)  intensity  of  stress. 

Let  /be  the  intensity  of  stress  tangential  to  the  ring. 
"     s  "     "  "          "      "       perpendicular  to  the  plane 

of  the  ring. 

Let  a-,  /?,  y  be  the  corresponding  strains. 

Let  E  and  mE  be  respectively  the  coefficients  of  direct  and 
lateral  elasticity. 

Then,  since  E,  f,  s  are  principal  stresses  (Chap.  IV), 


_  _ 

~ 


f     <?  +  *  f+g 


_  _  _ 

E        mE'          ~E    '   mE  '          ~  E  ~     mE  '    v 

But  y  is  constant.  Also,  since  the  ends  are  free,  the  total 
pressure  on  a  transverse  section  is  nil,  and  hence  it  might  be 
inferred  that  s  is  zero  at  every  point.  Adopting  this  value  of  s, 

By  eq.  (i), 

f-\-g=  a  constant  =  c  .......     (2) 

Again, 

d(qx)  —fdx  =  xdq  +  qdx  ......     (3) 

By  eqs.  (2)  and  (3), 

xdq  +  2qdx  =  cdx. 

.'.  d(x*q)  =  cxdx. 
Integrating, 

\    /       V      *V  =  f£  +  </,  .    ......    (4) 

c'  being  a  constant  of  integration. 

When  x  =  r,  the  internal  radius,  q  =/. 
"      j;  =  r;,  the  external  radius,  q  =pf. 


SPHERICAL    SHELLS.  591 

Hence,  by  eq.  (4), 

„_£  =  ,  =  ,.,_£; 

and  therefore 


Hence,  by  eq.  (4), 

r*p  —  r1 8/       p  —  P'     r'r' 


L'        «a  .J  a 


r  _  r  i  ;  •   •   •   (5) 

and,  by  eq.  (2), 

r*p  _  rf  *p'        p  —  p' 
f=  —* -7r-  +  ^-r 

J  ***  /*+'  *  I  >%* 


r'  —  r' '  x'      r  —  r' 

3.  Spherical  Shells. — Let  the  data  be  the  same  as  before. 
The  section  made  by  any  diametral  plane  must  develop  a  total 
resistance  of  2nrtf.  Then 

2nrtf  —  7tr*p, 
or 

*f  =  pr -     (0 

Hence,  a  spherical  shell  is  twice  as  strong  as  a  cylindrical 
shell  of  the  same  diameter  and  thickness  of  metal,  so  that  the 
strongest  parts  of  egg-ended  boilers  are  the  ends. 

Cor.  I.  Let  the  shell  be  subjected  to  an  external  pressure 
p'  as  well  as  to  an  internal  pressure  p.  Then 

2n  — —-tf=  nrp*  —  nr'  *#'. 


/"is  a  tension  or  a  pressure  according  as  r*p  ^  r'*p'. 

Generally,  r'  —  r  is  very  small,  and  the  relation  (2)  may  be 
written 

ft  =  r-(.P-p'} (3) 


592  THEORY  OF  STRUCTURES. 

Cor.  2.  For  a  thick  hollow  sphere,  Rankine  obtains 

P  ==  2Sr"  +  2r"  aPProximatelv-       •     •     •     (4) 

4.  Practical  Remarks. — A  common  rule  requires  that  the 
working  pressure  in  fresh-water  boilers  should  not  exceed  one- 
sixth  of  the  bursting  pressure,  and  in  the  case  of  marine  boilers 
that  it  should  not  exceed  one-seventh. 

An  English  Board  of  Trade  rule  is  that  the  tensile  working 
stress  in  the  boiler-plate  is  not  to  exceed  6000  Ibs.  per  square 
inch  of  gross  section,  and  French  law  fixes  this  limit  at  4250 
Ibs.  per  square  inch. 

The  thickness  to  be  given  to  the  wrought-iron  plates  of  a 
cylindrical  boiler  is,  According  to  French  law, 

/  =  .0036/2;- +  .1  in.; 
according  to  Prussian  law, 

/  —  (^•°°3«  _  i)r  +  i  in.  —  .oo$nr  +  .1  in.,  approximately, 

r  being  the  radius  in  inches,  and  n  the  excess  of  the  internal 
above  the  external  pressure  in  atmospheres. 

The  thickness  given  to  cast-iron  cylindrical  boiler-tubes  is, 
according  to  French  law,  five  times  the  thickness  of  equivalent 
wrought-iron  tubes ;  according  to  Prussian  law, 

t  _  ^.om  _  i)r  +  -J  in.  =  .01  nr  +  i  in.,  approximately. 

Steam-boilers  before  being  used  should  be  subjected  to  a 
hydrostatic  test  varying  from  \\  to  3  times  the  pressure  at 
which  they  are  to  be  worked. 

Fairbairn  conducted  an  extensive  series  of  experiments 
upon  the  collapsing  strength  of  riveted  plate-iron  flues,  by 
enclosing  the  flues  in  larger  cylinders  and  subjecting  them  to 
hydraulic  pressure.  From  these  experiments  he  deduced  the 
following  formula  for  a  ivrought-iron  cylindrical  flue  or  tube : 

Collapsing  pressure  I  =  ^  =403 1 50  ^ 

in  pounds  per  square  inch  of  surface  i  " 


PRACTICAL   RULES  FOR  BOILERS  AND   FLUES.          593 

t  "being  the  thickness  and  r  the  radius  in  inches,  and  /  the  length 
in  feet. 

This  formula  cannot  be  relied  upon  in  extreme  cases  and 
when  the  thickness  of  the  tube  is  less  than  f  in. 

Note. — In  practice,  t2  may  be  generally  used  instead  of  t2-19. 
The  experiments  also  showed  that  the  strength  of  an  elliptical 
tube  is  almost  the  same  as  that  of  a  circular  tube  of  which  the 
radius  is  the  radius  of  curvature  at  the  ends  of  the  minor  axis. 
Hence,  if  a  and  b  are  the  major  and  minor  axes  of  the  ellipse^ 
the  above  formula  becomes 

b  /2'19 
p  =  403150  -a  -j-. 

By  riveting  angle-  or  T-irons  around  a  tube,  its  length  is 
virtually  diminished  and  its  strength  is  therefore  increased,  as 
it  varies  inversely  as  the  length. 

The  thickness  of  tubes  subjected  to  external  pressure  is, 
according  to  French  law,  twice  the  thickness  of  tubes  subjected 
to  interior  pressure,  but  under  otherwise  similar  conditions  ; 
according  to  Prussian  law  the  thickness  of  heating  pipes  is 

/  =  .0067^  Vn  -f-  .05  in.,  if  of  sheet-iron, 
and 

t  —  .Old  Vn  +  .07  in.,  if  of  brass. 

According  to  Reuleaux,  the  thickness  (f)  of  a  round  flat 
plate  of  radius  r,  subjected  to  a  normal  pressure,  uniformly  dis- 
tributed and  of  intensity/,  is  given  by  the  formula 

7        t  _    A7 


according  as  the  plate  is  merely  supported  around  the  rim  or 
is  rigidly  fixed  around  the  rim,  as,  e.g.,  the  end  plates  of  a 
cylindrical  boiler ;  /",  as  before,  is  the  coefficient  of  strength. 
The  corresponding  deflections  of  the  plate  are 


-(-V 

6\t> 


and     -,   -  M=. 


594  THEORY  OF  STRUCTURES. 


EXAMPLES. 

1.  What  should  be  the  thickness  of  the  plates  of  a  cylindrical  boHer 
6  ft.  in  diameter  and  worked  to  a  pressure  of  50  Ibs.  per  square  inch,  in 
order  that  the  working  tensile  stress  may  not  exceed  1.67  tons  per  square 
inch  of  gross  section  ?  Ans.  .42  in. 

2.  A  cylindrical  boiler  with  hemispherical  ends  is  4  ft.   in  diameter 
and  22  ft.  in  length.    Determine  the  thickness  of  the  plates  for  a  steam- 
pressure  of  4  atmospheres. 

3.  What  is  the  collapsing  pressure  of  a  flue   10  ft.   long,  36  in.  in 
diameter,  and  composed  of  i-in.  plates?  Also  of  a  flue  30  ft.  long,  48  in. 
in  diameter,  and  T\  in.  thick?  Ans.  490.84  Ibs.;  91.59  Ibs. 

4.  Determine  the  thickness  of  a  2-in.  locomotive  fire-tube  to  support 
an  external  pressure  of  5  atmospheres. 

5.  A  copper  steam-pipe  is  4  in.  in  diameter  and  \  in.  thick.    Find  the 
working  pressure,  the  safe  coefficient  of   strength  for  copper  being  1000 
Ibs.  per  square  inch.  Ans.  125  Ibs.  per  square  inch. 

6.  A  /-ft.  boiler  of  ^-in.  plates  was  burst  at  a  longitudinal   double- 
riveted  joint  by  a  pressure  of  310  Ibs.  per  square  inch.     Find   the  coef- 
ficient of  ultimate  strength.  Ans.  29,760  Ibs. 

7.  A  50- in.  cylindrical  boiler  of  -^  in.   plates  is  made  of  wrought- 
iron  whose  safe  coefficient  of  strength  is  4000  Ibs.  per  square  inch.     Find 
the  working  pressure.  Ans.  50  Ibs.  per  square  inch. 

8.  A  lo-in.  cast-iron  water-pipe  is  subjected  to  a  pressure  of   250  Ibs. 
per  square  inch.     Find  its  thickness,  the  coefficient  of  working  strength 
being  2000  Ibs.  per  square  inch.  Ans.  i£  in. 

9.  A  steel  spherical  shell  36  in.  in  diameter  and  f  in.  thick  is  sub- 
jected to  an  internal  fluid  pressure  of  300  Ibs.  per  square  inch.     Find  its 
coefficient  of  strength.  Ans.  7200  Ibs. 

10.  A    thin,    hollow,    spherical,    elastic    envelope,    whose     internal 
radius  is  R,  was  subjected  to  a  fluid  pressure  which  caused  it  to  expand 
gradually  until  its  radius  became  R\  .     Determine  the  work  done. 

11.  The  plates  of   a  cylindrical  boiler  5  ft.  in  diameter  are  \  in.  thick. 
Find  to  what  pressure  the  boiler  may  be  worked  so  that  the  tensile  stress 
in  the  plates  may  not  exceed  i|  tons  per  square  inch  of  gross  section. 


EXAMPLES,  595 

12.  Show  that  the  assumption  of  a  uniform    distribution  of  stress  in 
the  thickness  of  a  cylindrical  or  spherical  boiler  is  only  admissible  when 
the  thickness  is  very  small. 

13.  A  metal  cylinder  of  internal  radius  r  and  external  radius   nr   is 
subjected  to  an  internal  pressure  of  p  tons  per  square  inch.     Show  that 
the   total    work   done   in   stretching  the   cylinder   circumferentially   is 

-~~  -j— -—  ft. -tons  per  square  foot  of  surface,  E  being  the  metal's  co- 
efficient of  elasticity. 

14.  The    cast-iron  cylinder  of  an  hydraulic  press  has   an  external 
diameter  twice  the  internal,  and  is  subjected  to  an  internal  pressure  of 
/tons  per  square  inch.  Find  the  pHncipal  stresses  at  the  outer  and  inner 
circumferences.  Also,  if  the  pressure  is  3  tons  per  square  inch,  and  if  the 
internal  diameter  is  10  in.,  find  the  work  done  in  stretching  the  cylinder 
circumferentially,  E  being  8000  Ibs. 

Ans.  At  inner  circumference,  q  =p,  a  thrust,  and/=  —  ^p,  a  tension. 
At  outer  circumference,  q  =  o,  and/  =  —  f /,  a  tension. 

Woik  =  126  ft.-lbs.  per  square  foot  of  surface. 

15.  The  chamber  of  a  2/-ton  breech-loader  has  an  external  diameter 
of  40  in.  and  an  internal  diameter  of  14  in.     Under  a  powder  pressure  of 
1 8  tons  per  square  inch,  find  the  principal  stresses  at  the  outer  and  inner 
circumferences,  and  also  the  work  done  ;  E  being  13,000  Ibs. 

Ans.  At  inner,  q  =  18  tons,  compression  ;  at  outer,  q  =  o. 

At  inner,/  =  —  23^  tons,  tension  ;  at  outer,/  =  —  5^¥  tons, 

tension. 
Work  =  \\  ft.-tons  per  sq.  ft.  of  surface. 

16.  What  should  be  the  thickness  of  a  o-in.  cylinder  (a)  which  has 
to  withstand  a  pressure  of  800  Ibs.  per  square  inch,  the  maximum  allow- 
able  tensile   stress  being  24,000  Ibs.  per  square  inch  ;   (ff)  which   has  to 
withstand  a  pressure  of  6000  Ibs.  per  square  inch  ;   the  maximum  allow- 
able tensile  stress  being  10,000  Ibs.  per  square  inch? 

Ans. — (a)  1.86  in.  ;  (b)  4^  in. 

17.  Show  that  the  radial  (or)  and  hoop  (fl)  strains  in  thick  hollow 

cylinders  and  spheres  are  connected  by  the  relation  a  —  — - — . 

1 8.  Prove  that  the  relation  in    Ex.  1.7  is  satisfied  by  the  values  ob- 
tained for/ and  q  in  the  Second  Method  of  Art.  2,  Chap.  X. 

19.  A  thick  hollow  sphere  of  internal  radius  r  and  external  radius 
nr  is  subjected  to  an  internal  pressure  p  and  an  external  pressure  p  . 
Determine  the  principal  stresses  at  a  distance  x  from  the  centre. 

_P'n*-p      p_-Jf    «V    .  p*?_-_p  _  p_-£_    n*S_ 

~   -  "          3        •          '    J  ~    •"  -« 


THEORY  OF  STRUCTURES. 

20.  Assuming  that  the  annulus  forming  the  section  of  a  cylindrical 
boiler  is  composed  of  a  number  of  infinitely  thin  rings,  show  that  the 

JL 

pressure  at  the   circumference  of  a  ring  of  radius  r  is  —     —  per  unit  of 

surface,  and  that  the  circumferential  stress  is  — h ,  A  and  B  de- 

r       mrm  + I 

noting  arbitrary  constants,  and  m  being  the  coefficient  of  lateral  con- 
traction. Find  the  values  of  A  and  B,  p*  and  pi  being  respectively  the 
internal  and  external  pressures. 

21.  Show  that  in  the  case  of  a  spherical  boiler  the  pressure  and  cir- 
cumferential stress  are  respectively and  —  +   .       Find 

'  ri+m  ri         (0»  —  l)f**  ~  3 

A  and  B. 

22.  Solve  Questions  i,  2,  6,  7,  8,  9,  and  n  on  the  supposition  that  /  is 
not  small  as  compared  with  r. 

23.  Taking/ =  4000  Ibs.  per  square   inch  and  E  =  30,000,000  Ibs., 
Find  the  thickness  and  deflection  of  the  end  plates  of  the  boiler  in  Ques- 
tion 7. 


CHAPTER   XL 
BRIDGES. 

1.  Classification. — Bridges  may  be  divided  into  four  gen- 
eral classes,  viz. :  (A)  Bridges  with  horizontal  girders ;  (B)  Can- 
tilever bridges  (Art.  15) ;  (C)  Suspension  bridges  (Chap.  XII); 
(D)  Arched  bridges  (Chap.  XIII).     The  present  chapter  treats 
of  bridges  in  Classes  A  and  B  only. 

2.  Comparative  Advantages  of  Curved  and  Horizontal 
Flanges  in  Girders  for  Bridges  of  Class  A. — The  depth  is 
sometimes  varied  for  the  sake  of   appearance,  and   it  is  also 
claimed  that,  an  economy  of  material  is  effected  by  giving  the 
chord  a  slope,  as,  e.g.,  in  the  case  of  the  Sault  Bridge  (Art.  19). 
Such  a  truss  is  intermediate  between  a  truss  with  horizontal 
flanges  and  one  of  the  parabolic  form.     The  curved  or  para- 
bolic form  is  not  well  suited  to  plate  construction,  and  a  dimi- 
nution in  depth  lessens  the  resistance  of  the  girder  to  distor- 
tion.    Again,  if  the  bottom  flange  is  curved,  the  bracing  for 
the  lower  part  of  the  girder  is  restricted  within   narrow  limits, 
and  the  girder  itself  must  be  independent,  so  that  in  a  bridge 
of  several  spans  any  advantage  which  might  be  derivable  from 
continuity  is  necessarily  lost.     Generally  speaking,  the  best  and 
most  economical  form  of  girder  is  that  in  which  the  depth  is 
uniform  throughout,  and  in  which  the  necessary  thickness  of 
flange  at  any  point  is  obtained  by  increasing  the  number  of 
plates. 

3.  Depth  of  Girder  or  Truss  (Class  A). — The  depth  usu- 
ally varies  from  one-fifteenth  to  one-seventh  (and  even  more)  of 
the  span.     It  is  generally  found  advisable  to  give  large  girders 

597 


598 


THEORY  OF  STRUCTURES. 


an  increased  depth,  and  they  should,  therefore,  be  designed  to 
have  a  specified  strength.  If  the  span  is  more  than  twelve  times 
the  depth,  the  deflection  becomes  a  serious  consideration,  and 
the  girder  should  be  designed  to  have  a  specified  stiffness.  The 
depth  should  not  be  more  than  about  i^  times  the  width  of 
the  bridge,  and  is  therefore  limited  to  24  ft.  for  a  single  and  to 
40  ft.  for  a  double-track  bridge. 

4.  Position  of  Platform. — The  platform  may  be  supported 
either  at  .ike-'top^r  bottom  flanges,  or  in  some  intermediate 
positidfu  In  favor/ oiVthe  last  it  is  claimed  that  the  main  girders 
may  be  braced  together  below  the  platform  (Fig.  365),  while 
the  upper  portions  s'tfrve  as  parapets  or  guards,  and  also  that 

,-the  vibration  communicated  by  a  passing  train  is  diminished. 

J.The  position,  however,  is  not  conducive  to  rigidity,  and  a  large 
amount  of  metal  is  required  to  form  the  connections. 


pi  A 


FIG.  365. 


FIG.  366. 


The  method  of  supporting  the  platform  on  the  top  flanges 
(Fig.  366)  renders  the  whole  depth  of  the  girder  available  for 
bracing,  and  is  best  adapted  to  girders  of  shallow  depth. 
Heavy  cross-girders  may  be  entirely  dispensed  with  in  the  case 
of  a  single-track  bridge,  and  the  load  most  effectively  distrib- 
uted, by  laying  the  rails  directly  upon  the  flanges  and  vertically 
above  the  neutral  line.  Provision  may  be  made  for  side  spaces 
by  employing  sufficiently  long  cross-girders,  or  by  means  of 
short  cantilevers  fixed  to  the  flanges,  the  advantage  of  the 


POSITION  OF  PLATFORM. 


599 


former  arrangement  being  that  it  increases  the  resistance  to 
lateral  flexure,  and  gives  the  platform  more  elasticity. 

Figs.  367,  368,  369  show  the  cross-girders  attached  to  the 
bottom  flanges,  and  the  desirability  of  this  mode  of  support 
increases  with  the  depth  of  the  main  girders,  of  which  the  cen- 
tres of  gravity  should  be  as  low  as  possible.  If  the  cross-girders 
are  suspended  by  hangers  or  bolts  below  the  flanges  (Fig.  369), 
the  depth,  and  therefore  the  resistance  to  flexure,  is  increased. 


FIG.  368. 


FIG.  369. 


In  order  to  stiffen  the  main  girders,  braces  and  verticals, 
consisting  of  angle-  or  tee  iron,  are  introduced  and  connected 
with  the  cross-girders  by  gusset  pieces,  etc. ;  also,  for  the  same 
purpose,  the  cross-girders  may  be  prolonged  on  each  side,  and 
the  end  joined  to  the  top  flanges  by  suitable  bars. 

When  the  depth  of  the  main  girders  is  more  than  about 
5  ft.,  the  top  flanges  should  be  braced  together.  But  the 
minimum  clear  headway  over  the  rails  is  16  ft.,  so  that  some 
other  method  should  be  adopted  for  the  support  of  the  plat- 
form when  the  depth  of  the  main  girders  is  more  than  5  ft. 
and  less  than  16  ft. 

Assume  that  the  depth  of  the  platform  below  the  flanges  is 
2  ft.,  and  that  the  depth  of  the  transverse  bracing  at  the  top  is 
i  ft.  ;  the  total  limiting  depths  are  7  ft.  and  19  ft.,  and  if  I  to  8 
is  taken  as  a  mean  ratio  of  the  depth  to  the  span,  the  corre- 
sponding limiting  spans  are  56  ft.  and  152  ft. 


600  THEORY   OF  STRUCTURES. 

5.  Comparative  Advantages  of  Two,  Three,  and  Four 
Main  Girders. — A  bridge  is  generally  constructed  with  two 
main  girders,  but  if  it  is  crossed  by  a  double  track  a  third  is 
occasionally  added,  and  sometimes  each  track  is  carried  by  two 
independent  girders. 

The  employment  of  four  independent  girders  possesses  the 
one  great  advantage  of  facilitating  the  maintenance  of  the 
bridge,  as  one-half  may  be  closed  for  repairs  without  interrupt- 
ing the  traffic.  On  the  other  hand,  the  rails  at  the  approaches 
must  deviate  from  the  main  lines  in  order  to  enter  the  bridge, 
so  that  the  width  of  the  bridge  is  much  increased,  and  far 
more  material  is  required  in  its  construction. 

Few,  if  any,  reasons  can  be  urged  in  favor  of  the  introduc- 
tion of  a  third  intermediate  girder,  since  it  presents  all  the 
objectionable  features  of  the  last  system  without  any  corre- 
sponding recommendation. 

The  two-girder  system  is  to  be  preferred,  as  the  rails,  by 
such  an  arrangement,  may  be  continued  over  the  bridge  with- 
out deviation  at  the  approaches,  and  a  large  amount  of  ma- 
terial is  economized,  even  taking  into  consideration  the  in- 
creased weight  of  long  cross-girders. 

6.  Bridge  Loads. — In  order  to  determine  the  stresses  in 
the  different  members  of  a  bridge  truss,  or  main  girder,  it  is 
necessary  to  ascertain  the  amount  and  character  of  the  load  to 
which  the  bridge  may  be  subjected.     The  load  is  partly  dead, 
partly  live,  and  depends  upon  the  type  of  truss,  the  span,  the 
number  of  tracks,  and  a  variety  of  other  conditions. 

The  dead  load  increases  with  the  span,  and  embraces  the 
weight  of  the  main  girders  (or  trusses),  cross-girders,  platform, 
rails,  ballast,  and  accumulations  of  snow. 

As  to  the  live  load  see  Art.  19. 

7.  Trellis    or    Lattice    Girders. — The  ordinary  trellis  or 
lattice  girder  consists  of  a  pair   of  horizontal  chords  and  two 
series  of  diagonals  inclined   in   opposite   directions  (Fig.  370). 
The  system  of  trellis  is  said  to  be  single,  double,  or  treble,  ac- 
cording to  the  number  of  diagonals  met  by  the  same  vertical 
section. 


TRELLIS   OR   LATTICE    GIRDERS. 


60 1 


Vertical  stiffeners,  united  to  the  chords  and  diagonals,  may 
be  introduced  at  regular  intervals. 


FIG.  370. 


371,  372,  373,  374  show  appropriate  sections  for  the 
top  chord  ;  the  bottom  chord  may  be  formed  of  fished  and 
riveted  plates,  or  of  links  and  pins. 


-ii  ir 


FIG.  371.  FIG.  372.  FIG.  373.  FIG.  374. 

The  verticals  and  diagonals  may  be  of  an  L,  T,  I,  H,  LJ  ,  or 
other  suitable  section,  but  the  diagonals,  except  in  the  case  of 
a  single  system  of  trellis,  are  usually  flat  bars,  riveted  together 
at  the  points  of  intersection. 

An  objection  to  this  class  of  girder  is  the  number  of  the 
joints. 

The  stresses  in  the  diagonals  are  determined  on  the  assump- 
tion that  the  shearing  force  at  any  vertical  section  is  equally 
distributed  between  the  diagonals  met  by  that  section,  which 
is  equivalent  to  the  substitution  of  a  mean  stress  for  the  differ- 
ent stresses  in  the  several  bars. 

E.g.,  let  w  be  the  permanent  load  concentrated  at  each  apex 
in  Fig.  370. 

Let  9  be  the  inclination  of  the  diagonals  to  the  vertical. 

The  reaction  at  A  —  fyw,  and  the  shearing  force  at  the 
section  MN  =  y^w  —  ^.w  =  2>2W- 

This  shearing  force  must  be  transmitted  through  the  diag- 
onals. 

Hence,  the  stress  in  ab  due  to  the  permanent  load 


—  sec 


-w  sec 


602  THEORY  OF  STRUCTURES. 

Again,  let  w'  be  the  live  load  concentrated  at  an  apex. 

The  greatest  shear  at  mn  due  to  the  live  load  occurs  when 
every  apex  between  a  and  7  is  loaded. 

This  shear  =  corresponding  reaction  at  I  =  -f-J-w',  and  the 
stress  in  ab  due  to  the  live  load 

=  i  X  -f- f  ™'  sec  0  —  f  f  w'  sec  8. 

Hence,  the  total  maximum  stress  in  ab  =  (|w  -f-  J J ze/)  sec  0. 

The  greatest  stress  of  a  kind  opposite  to  that  due  to  the 
dead  load  is  produced  in  ab  when  the  live  load  w'  is  concen- 
trated at  every  apex  between  I  and  b. 

The  shear  to  be  transmitted  is  then  2\w  due  to  the  dead  load, 
and  —  -j-fw'  due  to  the  live  load,  and  the  resultant  stress  in  ab 

sec  6  =  (f  w  —  l%w')  sec  6. 

This  stress  may  be  negative,  and  must  be  provided  for  by 
introducing  a  counter-brace  or  by  proportioning  the  bar  to 
bear  both  the  greatest  tensile  and  the  greatest  compressive 
stress  to  which  it  may  be  subjected. 

The  stress  in  any  other  bar  may  be  obtained  as  above. 

The  chord  stresses  are  greatest  when  the  live  load  covers 
the  whole  of  the  girder,  and  may  be  obtained  by  the  method 
of  moments,  or  in  the  manner  described  in  the  succeeding 
articles. 

In  the  above  it  is  assumed  that  the  members  of  the  girder 
are  riveted  together.  If  they  are  connected  by  pins,  each  of 
the  diagonal  systems  may  be  treated  as  being  independent. 

Thus,  the  system  I  2^^34567  transmits  to  the  supports 
the  stresses  due  to  loads  at  a,  3,  and  5. 

The  shear  due  to  the  dead  load,  transmitted  through  ab, 

3  w 

=  reaction  at  I  —  load  at  a  =  —w  —  w  —  —  . 

2  2 

W 

Hence,  the  stress  in  ab  due  to  the  dead  load  =  — -  sec  6. 

The  stress  in  ab  due  to  the  live  load  is  greatest  when  w'  is 
concentrated  at  each  of  the  points  3  and  5, 


WARREN  GIRDER. 


603 


The  maximum  shear  due  to  live  load  transmitted  through  ab 
=  ^wf  =  fee/, 

and  the  corresponding  stress  in  ab  =  £«/  sec  9. 
Hence,  the  total  maximum  stress  in  ab    • 


as  compared  with  (fcw  +  |- f  «/)  sec  0  obtained  on  the  first  as- 
sumption. 

8.  Warren  Girder. — The  Warren  girder  consists  of  two 
horizontal  chords  and  a  series  of  diagonal  braces  forming  a 
single  triangulation,  or  zigzag.  Fig.  375. 


N-1 


024  n  N-2 

FIG.  375. 

The  principles  which  regulate  the  construction  of  trellis 
girders  are  equally  applicable  to  those  of  the  Warren  type. 

The  cross-girders  (floor-beams)  are  spaced  so  as  to  occur  at 
the  apex  of  each  triangle. 

When  the  platform  is  supported  at  the  top  chords,  the  re- 
sistance of  the  structure  to  lateral  flexure  may  be  increased  by 
horizontal  bracing  between  the  cross-girders  and  by  diagonal 
bracing  between  the  main  girders. 

When  the  platform  is  supported  on  the  bottom  chords, 
additional  cross-girders  may  be  suspended  from  the  apices  in 
the  upper  chords,  which  also  have  the  effect  of  adding  to  the 
rigidity  of  the  main  girders. 

Let  w  be  the  dead  load  concentrated  at  an  apex  or  joint. 
«    „„.'  «     u    live       tt  u  "    "       *'       "       ** 

"    /      "     "    span  of  the  girder. 
"    k     "     "     depth  ".    " 
"   s      "    "    length  of  each  diagonal  brace. 
"    6     "    "    inclination  of  each  diagonal  brace  to  the  ver- 
tical. 
"   N -\-  i  be  the  number  of  joints. 


604  THEORY  OF   STRUCTURES. 

Then 


Two  cases  will  be  considered. 

CASE  I.  All  t  he  joints  loaded. 

Chord  Stresses.  —  These  stresses  are  greatest  when  the  live 
load  covers  the  whole  of  the  girder. 

Let  Sn  be  the  shearing  force  at  a  vertical  section  between 
the  joints  n  and  n  -f-  i. 

Let  Hn  be  the  horizontal  chord  stress  between  the  joints 
n  —  I  and  n  -f-  I- 

The  total  load  due  to  both  dead  and  live  loads 

=  (w-\-w')(N  -  i). 
The  reaction  at  each  abutment  due  to  this  total  load 


The  shearing  forces  in  the  different  bays  are 

UU  _!_  it)' 


w  A-  w' 
-~- 

Tfl  -J—  7£/ 


(N  —  i),  between  o  and  i  ; 
3),        „         !    .<     2. 


2     x     -  5),  2       3 ; 


and 


The  corresponding  diagonal  stresses  are 

50  sec  0,     .V,  sec  0,  SM  sec  0. 


WARREN   GIRDER.  605 

The  last  stresses  multiplied  by  sin  6  give  the  increments  of 
the  chord  stresses  at  each  joint.     Thus, 

Hl  =  tension  in  o  2  =  50  tan  0 

+  1  j  i  /    /      » r 

IV  /  IV  — f—  W   l>   1\   —    I 

fft  =  compression  in  i  3  =  ,S0  tan  0-|-  .S,  tan  0 
w  +  w'  IN—  i       w  -\-  w'  I  N  —  3 

_  W  +  tt/  /  2(7V  ~  2) 

-fiT3  =  tension  in  2  4  =  //,  +  5",  tan  0-\-  S9  tan  0 
-3) 


_  _ 

2  k          N 

—  compression  in  3  5  =  //!,  +  ^a  tan  ^  +  •$*  tan 

—  4) 


and   Hn  =  horizontal    stress    in    chord,    between    the    joints 

w  -\-  w'  /  n(N—n) 
n  —  i  and  n  +  i  —  —  !  --  T  -  X7       '  being   a    tension   for   a 

2t  K  J.\ 

bay  in  the  bottom  chord,  and  a  compression  for  a  bay  in  the 
top  chord. 

Note.—  The  same  results  may  be  obtained  by  the  method 
of  moments  ;  e.g.,  find  the  chord  stress  between  the  joints 
n  —  r  and  n  -|-  i. 

Let  a  vertical  plane  divide  the  girder  a  little  on  the  tight 
of  n. 

The  portion  of  the  girder  on  the  left  of  the  secant  plane  is 
kept  in  equilibrium  by  the  reaction  at  the  left  abutment,  the 
horizontal  stresses  in  the  chords,  and  the  stress  in  the  diagonal 
from  ;/  to  n  -\-  i. 

Take  moments  about  the  joint  n.     Then 


606  THEORY  OF  STRUCTURES. 


n  _      -  _     _ 

w-\-  w'    N  —  n 

~         2  N 

/.  Hn  =  etc. 

Diagonal  Stresses  due  to  Dead  Load. 

Let  dn  be  the  stress  in  the  diagonal  n,  n  -\-  I,  due  to  the 
dead  load. 

The  shearing  forces  in  the  different  bays  due  to  the  dead 
load  are 


w  w 

-(N  —  i),  between  o  and  I  ;  -(N  —  3),  between  I  and  2  ; 

2  2 


W  W 

-(^-5),      "       2    "    4;    -(N-j),  3    "    4; 


and  —  (N  —  2n  —  i),  between  n  and  n  -j-  I. 


The  corresponding  diagonal  stresses  are : 

w  w  s 

a  compression  -(N  —  i)  sec  6  =  -(N  —  i)-,  =  d^  inoi  ; 

—  2  A? 

a  tension  —(TV—  3)  sec  6  =  —(N  —  $)-r  =  dl  in  1 2 

2  2  A? 

a  compression  —(A7"—  5)  sec  0  —  -(N  —  5)7-  —  ^2  in  23  ; 

2  2  rt 

and  the  stress  in  the  nth  diagonal  between  n  and  n  +  i  is 


WARREN  GIRDER.  607 

being  a  tension  or  a  compression  according  as  the  brace  slopes 
down  or  up  towards  the  centre.  <i 

Diagonal  Stresses  due  to  Live  Load.  —  The  live  load  produces 
the  greatest  stress  in  any  diagonal  (n,  n  -\-  i),  of  the  same  kind 
as  that  due  to  the  dead  load,  when  it  covers  the  longer  of  the 
segments  into  which  the  diagonal  divides  the  girder.  Repre- 
sent this  maximum  stress  by  Dn. 

The  live  load  produces  the  greatest  stress  in  any  diagonal 
(n,  n  -|-  i),  of  a  kind  opposite  to  that  due  to  the  dead  load,  when 
it  covers  the  shorter  of  the  segments  into  which  the  diagonal 
divides  the  girder.  Represent  this  maximum  stress  by  Dnf. 

The  shearing  force  at  any  section  due  to  the  live  load,  as  it 
crosses  the  girder,  is  the  reaction  at  the  end  of  the  unloaded 
segment,  and  the  corresponding  diagonal  stress  is  the  product 

of  this  shearing  force  by  sec  6,  or  -r. 

The  values  of  the  different  diagonal  stresses  are  : 

£>„  =  compression  in  o  i  when  all  the  joints  are  loaded 

sw'N(N-  i) 
~~k^~      N~ 

Dl  =  tension  in  I  2  when  all  the  joints  except  one  are  loaded 

SW'(N- 


~  £7~  N        ~ 

^  =  compression  in  2  3  when  all  the  joints  except  I  and  2  are 


D^  =  tension  in  34  when  all  the  joints  except  I,  2,  and  3  are 

In^H          ^'(^-3)(^-4) 

loaded  =  J--       -Jr      -. 

Dn  =  stress  in  n,  n  -f-  i  when  all  the  joints  except  i,  2,  3,  .  .  . 

swf(N-n)(N-n-i) 
and  n  are  loaded  =  -  —-  ~v~~ 

DI  —  stress  in  o  I  before  the  load  comes  upon  the  girder  =  9. 


608  THEORY  OF  STRUCTURES. 

s  w' 
J}/=  compression  in  I  2  when  the  joint  I  is  loaded      =  -7  — i. 

K  IN 

/Y  =  tension  in  2  3  when  the  joints  I  and  2  are  loaded  =  -  —  3. 

£>,'=  compression  in  34  when  the  joints  I,  2,  and  3  are  loaded 

sw' 
=  kN6' 


JDn'=  stress  inn,n-{-i,  when  the  joints  I,  2,  .  .  .  and  ware  loaded 

s  w'  n(n-\-  i) 
=  'k~N~~~2~* 

The  total  maximum  stress  in  the  «th  diagonal  of  the  same 
kind  as  that  due  to  the  dead  load  =  dn  +  Dn  . 

The  resultant  stress  in  the  nth  diagonal  when  the  load 
covers  the  shorter  segment  =  dn  —  Dn'. 

This  resultant  stress  is  of  the  same  kind  as  that  due  to  the 
dead  load  so  long  as  dn  >  Dn',  and  need  not  be  considered  since 
dn  +  Dn  is  the  maximum  stress  of  that  kind. 

If  Dn'  >  dM1  it  is  necessary  to  provide  for  a  stress  in  the 
given  diagonal  of  a  kind  opposite  to  that  due  to  dn  -\-  DH,  and 
equal  in  amount  to  Dn'  —  dn. 

This  is  effected  by  counterbracing  or  by  proportioning  the 
bar  to  bear  both  the  stresses  dH  -)-  DH  and  Dn'  —  dn  . 

CASE  II.  Only  joints  denoted  by  even  numbers  loaded. 

0       2    '  4  N~2    N 


vvvvvv 

1       3  N-1 

FlG-  376- 
0       2       ^  N-2     N 


3  N-T 

FIG.  377- 


Chord  Stresses. — The  stresses  are  greatest  when  the  live  load 
covers  the  whole  of  the  girder. 


WARREN   GIRDER. 

The  total  load  due  to  both  dead  and  live  loads 


609 


The  reaction  at  each  abutment  due  to  this  total  load 


To  find  //i,  take  moments  about  I.     Then 

w  —  I—  w'  / 


To  find  H%,  take  moments  about  2. 


To  find  //g,  take  moments  about  3. 

uk  =  ™_?L(N  _  2)3  __  _  (w 

To  find  HI,  take  moments  about  4. 


w  4-  w'  „  / 

-—(7^-  2)4^  - 


To  find  Hn,  take  moments  about  n,  an 
Then 


let  ^^  be  even. 


and 


_ 

~ 


(A7'—  2)^      «(«  —  2} 
4~  4 

|-  w'  /  »(#"  ~  *) 


6  10  THEORY  OF  STRUCTURES. 

Next,  let  n  be  odd.     Then 


-  (w  +  w')x\(n  -  2)  +  (»  -  4)  +  •  •  •  +  5  +  3  + 

'  L  \  ^N  ~  2)n  _  (*  ~  I)' 

')~N\        4  4 

and 

wJrW'  I  ny —  2)n  —  (n  —  i)s 


N 


N 
Note.  —  If  —  is  even, 


ryy  _!_  ££/    I 

HN_,  the  stress  in  the  middle  bay,  =  -  g  —  -rtf. 

N  . 
If  —  is  odd, 

w  +  w'  I  N*  —  4 
HN_,  the  stress  in  the  middle  bay,  =  —  g  —  T  —  ^7  —  • 

Diagonal  Stresses  due  to  the  Dead  Load.  —  The  shearing  forces 

w 

in  the  different  bays  due  to  the  dead  load  are  —(N  '  —  2)  between 

4 

w  "W 

o  and  2,  -(N  —  6)  between  2  and  4,  -(TV—  10)  between  4  and 
4  4 

^,  etc. 

The  corresponding  diagonal  stresses  are 

s  w 

d0  in  o  i  =  T  -W  —  2)    =  dl  in  I  2  ; 
£  4  v 


i,m  23  =(N~  4)    =4  in  34; 


/4  in  4  5  =  j  ^-  10)  n=  db  in  56; 

*  4 

etc.,  etc.,  etc. 


HOWE    TRUSS. 


611 


Thus  the  stresses  in  the  diagonals  which  meet  at  an  unloaded 
joint  are  equal  in  magnitude  but  opposite  in  kind. 

Diagonal  Stresses  due  to  the  Live  Load. — These  are  found 
as  in  Case  I,  and 


-A, 


N 


If  —  is  odd,  there  is  a  single  stress  at  the  foot  of  each  of 

these  columns. 

The  maximum  resultant  stress  due  to  both  dead  and  live 
loads  is  obtained  as  before. 

E.g.,  the  maximum  resultant  stress  in  3  4  when  the  longer 
segment  is  loaded 


and  the  maximum  resultant  stress  in  3  4  when  the  shorter  seg- 
ment is  loaded 

=  4-  D;  =  A  -  A'- 

Note.  —  6  is  generally  60°,  in  which  case  s  —  2  ^.. 

9.  Howe  Truss.  —  Fig.  378  is  a  skeleton   diagram   of   a 
Howe  truss. 


FIG.  378. 

The  truss  may  be  of  timber,  of  iron,  or  of  timber  and  iron 
combined. 

The  chords  of  a  timber  truss  usually  consist  of  three  or 
more  parallel  members,  placed  a  little  distance  apart  so  as  to 
allow  iron  suspenders  with  screwed  ends  to  pass  between  them 
(Figs.  379  and  380). 


6l2 


THEORY  OF  STRUCTURES. 


Each  member  is  made  up  of  a  number  of  lengths  scarfed 
or  fished  together  (Figs.  381  and  382). 

The  main  braces,  shown  by  the  full  diagonal  lines  in  Fig. 
378,  are  composed  of  two  or  more  members. 

The  counter-braces,  which  are  introduced  to  withstand  the 
effect  of  a  live  load,  and  are  shown  by  the  dotted  diagonal 
lines  in  Fig.  378,  are  either  single  or  are  composed  of  two  or 
more  members.  They  are  set  between  the  main  braces,  and 
are  bolted  to  the  latter  at  the  points  of  intersection. 

The  main  braces  and  counters  abut  against  solid  hard-wood 
or  hollow  cast-iron  angle-blocks  (Fig.  380).  They  are  designed 
to  withstand  compressive  forces  only,  and  are  kept  in  place  by 
tightening  up  the  nuts  at  the  heads  of  the  suspenders. 

At 


FIG.  380. 


FIG.  381. 


FIG.  382. 


FIG.  383. 


FIG.  384. 


The  angle-blocks  extend  over  the  whole  width  of  the  chords ; 
if  they  are  made  of  iron,  they  may  be  strengthened  by  ribs. 

If  the  bottom  chord  is  of  iron,  it  may  be  constructed  on  the 
same  principles  as  those  employed  for  other  iron  girders.  It 
often  consists  of  a  number  of  links,  set  on  edge,  and  connected 
by  pins  (Figs.  383  and  384).  In  such  a  case  the  lower  angle- 
blocks  should  have  grooves  to  receive  the  bars,  so  as  to  prevent 
lateral  flexure. 

If  the  truss  is  made  entirely  of  iron,  the  top  chord  may  be 
formed  of  lengths  of  cast-iron  provided  with  suitable  flanges 
by  which  they  can  be  bolted  together.  Angle-blocks  may  also 
be  cast  in  the  same  piece  with  the  chord. 

To  determine   the   stresses   in    the   different   members,  the 


HOWE    TRUSS.  613 

same  data  are  assumed  as  for  the  Warren  girder,  except  that 
N  is  now  the  number  of  panels. 

Chord  Stresses.  —  These  stresses  are  greatest  when  the  live 
load  covers  the  whole  of  the  girder. 

Let  Hn  be  the  chord  stress  in  the  nth  panel. 

The  total  load  due  to  both  dead  and  live  loads 

=  (w  +  w'}(N  —  i). 
The  reaction  at  each  abutment  due  to  this  total  load 


Let  a  plane  MM'  divide  the  truss  as  in  Fig.  378.  The  por- 
tion of  the  truss  on  the  left  of  the  secant  plane  is  kept  in 
equilibrium  by  the  load  upon  that  portion,  the  reaction  at  the 
left  abutment,  the  chord  stresses  in  the  nth  panels,  and  the 
tension  in  the  nth  suspender. 

First,  let  the  load  be  on  the  top  chord  and  take  moments 
about  the  foot  of  the  nth  suspender.  Then 


or 


w  -f-  w'  n(N  —  n) 


_  w  +  w'  I  n(N—  ri) 

ff-  ~~~~' 


Next,  let    the   load    be    on    the   bottom    chord    and  take 
moments  about  the  head  of  the  nth  suspender.     Then 

w  +  w'    n(N  —  n) 
Hnk  —  -  L  —  /  --  —  —  -,  as  before. 

Thus,  Hn  is  the   same   for  corresponding  panels,  whether 
the  load  is  on  the  top  or  bottom  chord. 

Diagonal  Stresses  due  to  the  Dead  Load.  —  Let  VMf  be  the 


6l4  THEORY  OF  STRUCTURES. 

shearing  force  in  the  nth  panel,  or  the  tension  on  the  nth  sus- 
pender due  to  the  dead  load. 

First,  let  the  load  be  on  the  top  chord.     Then 

w  (N  —  i         \ 

VH'  =  —  (N  —  i)  —  nw  =  w\  -  •  —  n\. 

Next,  let  the  load  be  on  the  bottom  chord.     Then 

lit  //V-U  I  \ 

Vm'  =  --(N  _,)_(«_  i)w  =  w(—t-  -  n]. 
The  corresponding  diagonal  stresses  are 


and 

N+i 


Diagonal  Stresses  due  to  the  Live  Load. — Let  Vn"  be  the 
shearing  force  in  the  nth  panel,  or  tension  on  the  nth  suspen- 
der, when  the  live  load  covers  the  longer  segment. 

First,  let  the  load  be  on  the  top  chord. 

The  greatest  stress  in  the  nth  brace,  of  the  same  kind  as  that 
produced  by  the  dead  load,  occurs  when  all  the  panel  points 
on  the  right  of  MM'  are  loaded.  With  such  load,  Vn",  the 
shearing  force  on  the  left  of  MM ',  =  the  reaction  at  o 

«/'  N-n 

—  --(7VT  _  n  —  i) — TT— , 
2^  N 

and  the  corresponding  diagonal  stress,  Dn , 


___ 
-  k  2  (1V  I}     N 


Hence,  the  resultant  tension  on  the  nth  suspender  due  to 
both  dead  and  live  loads  =  Vn  =  Vnr  +  Vn"  . 


N-  \         \       w' 

-  -  n    +  ~(N-  n  - 


HOWE    TRUSS.  615 

and  the  resultant  maximum  compression  on  the  nth  brace  due 
to  both  dead  and  live  loads 

V  N-n 


The  live  load  tends  to  produce  the  greatest  stress  in  the 
nth  counter  when  it  covers  the  shorter  segment  up  to  and  in- 
cluding the  nth  panel  point.  Even  then  there  will  be  no  stress 
in  the  counter  unless  the  effect  of  the  live  load  exceeds  that 
of  the  dead  load  in  the  (n  +  i)th  brace. 

The  shearing  force  on  the  right  of  MM'  =  the  reaction  at  N 

_  w'  n(n  -\-  i) 


Hence, 

ZV»  tne  corresponding  diagonal  stress,  =  -  -  7T     , 

w     2  M  V 

and  the  resultant  stress  in  the  counter  =  Dnf  —  dn+t 
s  (  w'  n(n  +  i)  JV- 


Next,  let  the  load  be  on  the  bottom  chord.     Then 


and 

D   =  SW-(N      riN-n+l 
k   2  ^  N 

Hence, 


and 


Also,  the  stress  in  the  ^th  counter  is 

N  —  i 


6i6 


THEORY  OF  STRUCTURES. 


.—K  common  value  of  0  is  45°,  when  sec  6  =     =  1.414, 


and  tan  6  =  -rr/,  —  l- 
NK 

The  end  panels  and  posts,  shown  by  the  dotted  lines  in 
Fig.  378,  may  be  omitted  when  the  platform  is  suspended  from 
the  lower  chords. 

10.  Single  and  Double  Intersection  Trusses.  —  Fig. 
385  represents  the  simplest  form  of  single-intersection  (or 


FIG.  385. 

Pratt)  truss  ;  i.e.,  a  truss  in  which  a  diagonal  crosses  one  panel 
only.  It  may  be  constructed  entirely  of  iron  or  steel,  or  may 
have  the  chords  and  verticals  of  wood.  The  verticals  are  in 
compression  and  the  diagonals  in  tension.  The  angle-blocks 
are  therefore  placed  above  the  top  and  below  the  bottom 
chord.  Counter-braces,  shown  by  the  dotted  diagonals,  are  in- 
troduced to  withstand  the  effect  of  a  live  load. 

If  the  truss  is  inverted  it  becomes  one  of  the  Howe  type, 
and  the  stresses  in  the  several  members  of  both  trusses  may 
be  found  in  precisely  the  same  manner. 

Fig.    386    represents    a    double-intersection    (or   Whipple) 


XXXXXXX 


/ 


7 
FIG.  386. 


truss,  i.e.,  a  truss  in  which  a  diagonal  crosses  two  panels.  It 
may  be  constructed  entirely  of  iron  or  steel.  It  is  of  the  pin- 
connected  type,  and  the  two  diagonal  systems  may  be  treated 
independently. 

Let  0'  be  the  inclination  of  AB  to  the  vertical. 
"     *    "     "  "  Ai,  CD,  .  ..  to  the  vertical. 

Chord  Stresses.  —  These  stresses  are  greatest  when  the  live 
load  covers  the  whole  of  the  girder. 

The  reaction  at  A  from  the  system  A  BCD  .  .  .  =  4(wjrw')  ; 

"       "  A    "     «•      -     4*23...  =  («H-*»'l 


SINGLE  AND  DOUBLE   INTERSECTION   TRUSSES.       6l/ 

Wj  w'  being  the  dead  and  live  loads  concentrated  at  the  panel 
points  C,  2,  E,  4,  .  .  . 

The  shearing  forces  in  the  different  bays  are  : 

4(w  +  w')  in  AC,  from  the  system  ABCD  .  .  . 
(w  +  w')  in  AC,     "       "        "        ^123... 


3w       w      n  <2, 

|(w  _|_  w')  in  2E,      "        "        "        ^4  I  2  3 
2(w  +  «/')  in  £4,      "        "        " 
KW  +  w')  in  46,      "        "        " 
i(w  +  z£/r)  in  G6,      "        "        c< 
j(«,  _j.  «,')  in  67,        •"        "        "       ^123... 
The  corresponding  diagonal  stresses  are  : 
4(w  +  w'}  sec  6^  in  ^.5  ;  $%(w  +  w')  sec  6  in  ^4  I  ; 

3(0;  -f  w/)  sec  ^  in  ^^  J     2i(w  +  w')  sec  ^  in  23  J     etc- 
Hence,  the  /<?/  chord  stresses  are  : 

£  in  AC  =  4w  +  w')  tan  0r  +  3^(0;  +  *£/)  tan  6^; 

Ca  in  C2=C,+  3(w  +  w')  tan  0 

=  4(0;  +  w')  tan  0'  +  6J(w  +  w'}  tan  0  ; 

C,  in  2^  =  (T2  +  2\(w  -\-  w')  tan  0 

=  4(w  +  wr)  tan  6'  +  9(0;  +  w')  tan  6^  ;     etc. 

The  bottom  chord  stresses  are  : 
7;  in  j5i  =  4(w  +  w')  tan  ^  ; 

T9  in  iZ?  =  T;  +  3i(w  +  w/)  tan  e 

=  4(w  +  w')  tan  07  +  3i(w  +  «/)  tan  0=C,. 

So,        TZ  =  C,,     T4  =  C3,      etc.,  etc. 

Again,  the  stress  in  any  diagonal  4  5  of  the  system  A  I  2  .  .  . 
due  to  the  dead  load  =  \\w  sec  0. 

The  live  load  produces  the  greatest  stress  in  4  5,  of  the  same 


6l8  THEORY  OF  STRUCTURES. 

kind  as  that  due  to  the  dead  load,  when  it  is  concentrated  at 
all  panel  points  of  the  system  A  I  2  3  ...  on  the  right  of  4. 

The  reaction  at  A  is  then  ^w' ,  and  the  corresponding 
diagonal  stress  =  i£w'  sec  0. 

Hence  the  maximum  resultant  stress  in  45  =  (f-«> -f- J/ze/) 
sec  ft 

The  live  load  tends  to  produce  the  greatest  stress  in  any 
counter  5  8  when  it  is  concentrated  at  all  the  panel  points  of 
the  system  A  I  2  3  ...  on  the  left  of  8. 

The  reaction  at  the  right  abutment  is  then  £«/,  and  the 
corresponding  stress  in  the  counter  =  £ze/  sec  0.  Thus,  the 
resultant  stress  in  the  counter  =  (f  w' —  %w)  sec  0,  \w  sec  6  being 
the  stress  in  6  7  due  to  the  dead  load. 

Similarly,  the  stresses  in  any  other  diagonal  and  counter 
may  be  found. 

The  Pratt  truss  composed  entirely  of  iron  and  with  some 
of  the  details  of  the  Whipple  truss  is  sometimes  called  a 
Murphy-Whipple  truss.  The  Linville  truss  is  a  Whipple  truss 
made  of  wrought-iron,  the  verticals  being  tubular  columns. 

II.  Post  and  Quadrangular  Trusses. — The  peculiarity  of 
the  Post  truss  (Fig.  387)  is  that  the 
struts   are    inclined    at    an    angle    of 
about  22°  30'  to  the  vertical,  with  a 
FlG  387-  view    to    an    economy    of    material. 

The  ties  cross  two  panels  at  an  angle  of  45°  with  the  vertical. 

In   the    quadrangular    truss 
(Fig.  388)  the  bottom  chord  has 


additional  points  of  support  half-  FIG.  388. 

way  between  the  panel  points. 

The  Bollman,  Fink,  and  other  bridge-trusses  have  been 
referred  to  in  a  previous  chapter. 

12.  Bowstring  Girder  or  Truss. — The  bowstring  girder 
in  its  simplest  form  is  represented  by  Fig.  389,  and  is  an  excel- 
lent structure  in  point  of  strength  and  economy. 

The  top  chord  is  curved,  and  either  springs  from  shoes 
(sockets)  which  are  held  together  by  a  horizontal  tie,  or  has  its 
ends  riveted  to  those  of  the  tie. 

The  strongest  bow  is  one  composed  of  iron  or  steel  cylin- 


BOWSTRING    TRUSS. 


619 


drical  tubes,  but  any  suitable  section  may  be  adopted,  and  the 
inverted  trough  offers  special  facilities  for  the  attachment  of 
verticals  and  diagonals. 

The  tie  is  constructed  on  the  same  principles  as  those  em- 
ployed for  other  iron  girders,  but  in  its  best  form  it  consists  of 
flat  bars  set  on  edge  and  connected  with  the  shoes  by  gibs  and 
cotters. 


The  platform  is  suspended  from  the  bow  by  means  of  ver- 
tical bars  which  are  usually  of  an  I  section,  and  are  set  with 
the  greatest  breadth  transverse,  so  as  to  increase  the  resistance 
to  lateral  flexure.  In  large  bridges  the  webs  of  verticals  and 
diagonals  may  be  lattice-work. 

If  the  load  upon  the  girder  is  uniformly  distributed  and 
stationary,  verticals  only  are  required  for  its  suspension,  and 
the  neutral  axis  of  the  bow  should  be  a  parabola.  An  irregu- 
larly distributed  load,  such  as  that  due  to  a  passing  train,  tends 
to  change  the  shape  of  the  bow,  and  diagonals  are  introduced 
to  resist  this  tendency. 

A  circular  arc  is  often  used  instead  of  a  parabola. 

To  determine  the  stresses  in  the  different  members,  assum- 
ing that  the  axis  ABC  of  the  top  chord  is  a  parabola : 

Let  w  be  the  dead  load  per  lineal  foot. 
"     w'  "     "     live      " 
"     /     "     "     span  of  the  girder. 
"     k    "     "     greatest  depth  BD  of  the  girder. 

Chord  Stresses. — These  stresses  are  greatest  when  the  live 
load  covers  the  whole  of  the  girder. 

The  total  load  due  to  both  dead  and  live  loads  —  (w  -j-  w')l. 
The  reaction  at  each  abutment  due  to  this  total  load 


/. 


620 


THEORY  OF  STRUCTURES. 


Let  H  be  the  horizontal  thrust  at  the  crown. 
«     T  "      "  "  tension  in  the  tie. 

Imagine  the  girder  to  be  cut  by  a  vertical  plane  a  little  on 
the  right  of  BD.  The  portion  ABD  is  kept  in  equilibrium  by 
the  reaction  at  A,  the  weight  upon  AD,  and  the  forces  H 
and  T. 

Take  moments  about  B  and  D.     Then 


Tk  = 


and 


=  Hk, 


Let  H'  be  the  thrust  along  the  chord  at  any  point  P. 

Let  x  be  the  horizontal  distance  of  P  from  B. 

The  portion  PB  is  kept  in  equilibrium  by  the  thrust  H  at 
B,  the  thrust  H'  at  P,  and  the  weight  (w  +  w')x  between  P 
and  B.  Hence, 

H*  sec'  i  =  H'*  =  H*  +  (w  +  ^v}*x\ 

i  being  the  inclination  of  the  tangent  at  P  to  the  horizontal, 
and 

tw  +  w'\t  r 

the  thrust  at  A  =  —  + 


Diagonal  Stresses  due  to  Live  Load.  —  Assume  that  the  load  is 
concentrated  at  the  panel  points,  and  let  it  move  from  A 
towards  C. 


If  the  diagonals  slope  as  in  Fig.  390,  they  are  all  ties,  and 
the  live  load  produces  the  greatest  stress  in  any  one  of  them, 
as  QS,  when  all  the  panel  points  from  A  up  to  and  including 
Q  are  loaded. 


BOWSTRING    TRUSS.  621 

Let  ;r,  y  be  the  horizontal  and  vertical  co-ordinates,  respec- 
tively, of  any  point  on  the  parabola  with  respect  to  B  as 
origin. 

The  equation  of  the  parabola  is 


Let  the  tangent  at  the  apex  P  meet  DB  produced  in  L, 
and  DC  produced  in  E. 

Draw  the  horizontal  line  PM. 

From  the  properties  of  the  parabola,  LM  =  2BM. 

Let  PM  —  x  and  BM  =  y. 

From  the  similar  triangles  LMP  and  LDE, 


.- 

~'~~ 


x+QE' 


A,,»,     „ 

.  &L  _  l~2x 

''  QE~  1+2X' 

Draw  EF  perpendicular  to  QS  produced,  and  imagine  the 
girder  to  be  cut  by  a  vertical  plane  a  little  on  the  right  of  PQ. 

The  portion  of  the  girder  between  PQ  and  C  is  kept  in 
equilibrium  by  the  reaction  R  at  C>  the  thrust  in  the  bow  at  P, 
the  tension  in  the  tie  at  Q,  and  the  stress  in  the  diagonal  QS. 

Denote  the  stress  by  Dn  ,  and  let  the  panel  OQ  be  the  nth. 

Let  0  be  the  inclination  of  QS  to  the  horizontal. 

Take  moments  about  E.     Then 


RX  CE, 
or 

Dn  =  R*      cosec  B  .......     (3) 


622  THEORY  OF  STRUCTURES. 

Let  N  be  the  total  nuntber  of  panels.     Then 

is  a  panel  length,    and    w'-r=  is  a  panel  weight. 


Also,  x  =  njj  —  -,   and  hence 


l-2x  _CE___  N—n 
7+  2x  ~  QE  ~~     ~~^~* 

R,  the  reaction  at  C  when  the  n  panel  points  preceding  T 
are  loaded, 

n(n 

Thus,  equation  (3)  becomes 

A-y/(»+i)^^cosecft    ....    (4) 
Again,  by  equation  (i), 


_ 

"  2         ^2 


N* 

and 


ST'  ST 

Hence,  finally, 

/    H -\TI  \  (•*»  —  #  —  I )(^  ~T~  I ) I 

•M  AT 2    <  v  /\ 

±-    (5) 


This  formula  evidently  applies  to  all  the  diagonals  between  D 
and  C. 


BOWSTRING    TRUSS.  623 

Similarly,  it  may  be  easily  shown  that  the  stress  in  any 
diagonal  between  D  and  A  is  given  by  an  expression  of  pre- 
cisely the  same  form. 

Hence,  the  value  of  Dn  in  equation  (5)  is  general  for  the 
whole  girder. 

A  load  moving  from  C  towards  A  requires  diagonals  in- 
clined in  an  opposite  direction  to  those  shown  in  Fig.  390. 

Stresses  in  the  Verticals  due  to  the  Live  Load. — Let  VH  be  the 
stress  in  the  ?zth  vertical  PQ  due  to  the  live  load.  This  stress 
is  evidently  a  compression,  and  is  a  maximum  when  all  the 
panel  points  from  A  up  to  and  including  O  are  loaded. 

Imagine  the  girder  to  be  cut  by  a  plane  S' S"  very  near  PO, 
Fig.  390.  The  portion  of  the  girder  between  S'S"  and  C  is  kept 
in  equilibrium  by  the  reaction  R'  at  C,  the  thrust  in  the  bow 
at  P,  the  tension  in  the  tie  at  O,  and  the  compression  Vn  in 
the  vertical. 

Take  moments  about  E.     Then 

VnQE  =  X'X  CE,     or     Vn  =  R'^^-- , 


and  R',  the  reaction  at  C  when  the  (n  —  i)  panel  points  from  A 
to  and  i 
Hence, 


up  to  and  including  O  are  loaded,  =  — /  - 


Vn  =  —  l- 

2 


a  general  formula  for  all  the  verticals. 

Let  vn  be  the  tension  in  the  nth  vertical  due  to  the  dead 
load.  The  resultant  stress  in  it  when  the  live  load  covers  AO 
is  v^r-  VH,  and  if  negative,  this  is  the  maximum  compression 
to  which  PQ  is  subjected. 

\ivn  —  Vn  is  positive,  the  vertical  PQ  is  never  in  compression. 

The  maximum  tension  in  a  vertical  occurs  when  the  live 

load  covers  the  whole  of  the  girder  and  =  w'--*j  +  the  tension 

due  to  the  dead  load. 

Note. — The  same  results  are  obtained  when  N  is  odd. 


624  THEORY  OF  STRUCTURES. 

13.  Bowstring  Girder  with  Isosceles  Bracing. 

Diagonal  Stresses  due  to  the  Dead  Load. — Under  a  dead  load 
the  bow  is  equilibrated  and  the  tie  is  subjected  to  a  uniform 
tensile  stress  equal  in  amount  to  the  horizontal  thrust  at  the 
crown.  The  braces  merely  serve  to  transmit  the  load  to  the 
bow  and  are  all  ties. 

Let  Tl ,  7",  be  the  tensile  stresses  in  the  two  diagonals 
meeting  at  any  panel  point  Q.  Let  0, ,  #2  be  the  inclinations 
of  the  diagonals  to  the  horizontal. 

Let  W  be  the  panel  weight  suspended  from  Q. 


The  stress  in  the  tie  on  each  side  of  Q  is  the  same,  and 
therefore  Tl ,  7^,  and  Ware  necessarily  in  equilibrium. 
Hence, 


T*  1X7  I  J  T*  117 

Tt  —  W - — -TT. — ;— 77-7,     and      Tz  =  W- 

Sin  \y j  -J-  c/2)  Sin  ^i/j  — p  t/2y 

Diagonal  Stresses  due  to  the  Live  Load. — Let  TV  be  the  num- 
ber of  half  panels. 

2l 
The  length  of  a  panel  =  -ry  ;  the  weight  at  a  panel  point 


Let  the  load  move  from  A  towards  C.  All  the  braces  in- 
clined like  OP  are  ties,  and  all  those  inclined  like  QP  are  struts. 

The  live  load  produces  the  greatest  stress  in  OP  when  it 
covers  the  girder  between  A  and  O. 

Denote  this  stress  by  Dn ;  OG  is  the  «th  half-panel. 

As  before, 

RX  CE (i) 


BOWSTRING   GIRDER    WITH  ISOSCELES  BRACING.      62$ 

I  w'nl(n-\-2] 

The  load  upon  AO  =•  nw  -j-=,  and  hence  R  =  -j~r'  -  TT—  . 

The  ratio  of  CE  to  EF  is  denoted  by  the  same  expression 
as  in  the  preceding  article.     Thus, 

«  _i- 

_w'ln-\-2N—  n[.          N 
"==  ~8~;&»+i      AT  N-n-  I  ~'.'        ' 

The  live  load  produces  the  greatest  stress  in  OM  when  it 
covers  the  girder  up  to  and  including  D. 

Denote  the  stress  by  Dnf  ;  DG  is  now  the  nth  half  panel. 
Let  Rf  be  the  reaction  at  C. 
As  before, 

CE 

cosec  6,  ......    (3) 


0  being  tjae  angle  MOD. 

The  weight  upon  AD  =  (n  —  i)ze/-r~ 
and  hence 


2         A"'    ' 

It  may  be  easily  shown,  as  in  the  preceding  article,  that 


CE      N-n-,       A  . 

TvS  —  -  i  -  y  and  cosec  0  =  N  -  TT^TT  -  - 
OE          n+i  4nk(N—n) 


7"")    ^  "^  f     \ 

n  ~~~~~%k      n  N  N-n  ~ 

Hence,  when  the  load  moves  from  A  towards  C,  eq.  (2) 
gives  the  diagonal  stress  when  n  is  even,  and  eq.  (4)  gives  the 
stress  when  n  is  odd. 

If  the  load  moves  from  C  towards  A,  the  stresses  are  re- 
versed in  kind,  so  that  the  braces  have  to  be  designed  to  act 
both  as  struts  and  ties. 


626 


THEORY  OF  STRUCTURES. 


. — By  inverting  Fig.  391,  a  bowstring  girder  is  obtained 
with  the  horizontal  chord  in  compression  and  the  bow  in 
tension. 

14.  Bowstring  Suspension  Bridge  (Lenticular  Truss}. — 
This  bridge  is  a  combination  of  the  ordinary  and  inverted 
bowstrings.  The  most  important  example  is  that  erected  at 
Saltash,  Cornwall,  which  has  a  clear  span  of  445  feet.  The 
bow  is  a  wrought-iron  tube  of  an  elliptical  section  stiffened 
at  intervals  by  diaphragms,  and  the  tie  is  a  pair  of  chains. 

A  girder  of  this  class  may  be  made  to  resist  the  action  of  a 
passing  load  either  by  the  stiffness  of  the  bow  or  by  diagonal 
bracing. 

B     Q 


In  Fig.  392,  let  BD  =  k,  B'D  =  k'. 

Let  //be  the  horizontal  thrust  at  B,  and  T  the  horizontal  pull 
at  B' ',  when  the  live  load  covers  the  whole  of  the  girder.     Then 

fjei  _] not'  72 

•  rr1 

L     i      z./  —    •*• 


8 
First,  iet  k  =  k' .     Then 

TT y 


f 


1  6 


which  is  one  half  of  the  corresponding  stress  in  a  bowstring 
girder  of  span  /  and  depth  k. 

One  half  of  the  total  load  is  supported  by  the  bow  and  one 
half  is  transmitted  through  the  verticals  to  the  tie.     Hence, 

the  stress  in  each  vertical  —  -(w'  -\-  w"\ 


w"  being  the  portion  of  the  dead  weight  per  lineal  foot  borne 
by  the  verticals,  and  A^the  number  of  panels. 


CANTILEVER    TRUSSES.  627 

The  diagonals  are  strained  only  under  a  passing  load. 

Let  PP'  be  a  vertical  through  £,  the  point  of  intersection 
of  any  two  diagonals  in  the  same  panel,  and  let  the  load  move 
from  A  towards  O. 

By  drawing  the  tangent  at  P  and  proceeding  as  in.  Art.  13, 
the  expression  for  the  diagonal  stress  in  QS  becomes,  as  before, 


Similarly,  the  stress  in  the  vertical  QQ  is 
„  /       w'n(n  —  i)l  - 


Next,  let  k  and  k'  be  unequal. 

Let  Wbe  the  weight  of  the  bow,  W  the  weight  of  the  tie. 

Then,  under  these  loads, 

~%"k  =  H^H'     :~8~£'    °r    W'~~k'*       '     '     ^ 

The  verticals  are  not  strained  unless  the  platform  is  attached 
to  them  along  the  common  chord  ADO.  In  such  a  case,  the 
weight  of  the  platform  is  to  be  included  in  W' 

The  tangents  at  P  and  P'  evidently  meet  AO  produced  in 
the  same  point  O ',  for  EO'  is  independent  of  k  or  k'.  Hence, 
the  stresses  in  the  verticals  and  diagonals  due  to  the  passing 
load  may  be  obtained  as  before. 

15.  Cantilever  Trusses. — A  cantilever  is  a  structure  sup- 
ported at  one  end  only,  and  a  bridge  of  which  such  a  structure 
forms  part  may  be  called  a  cantilever  bridge.  Two  cantilevers 


BEUD.GE.OVER  ST.  LAWRENCE  AT  NIAGARA. 
FIG.  393. 


may  project  from  the  supports  so  as  to  meet,  or  a  gap  may  be 
left  between  them   which  may  be  bridged  by  an  independent 


628 


THEORY  OF  STRUCTURES. 


girder  resting  upon  or  hinged  to  the  ends  of  the  cantilevers. 
The  form  of  the  cantilever  is  subject  to  considerable  variation. 


SUKKUR  BRIDGE 
FIG.  394. 


\  /  \  /^T*"^^*«^_ 


FORTH  BRIDGE. 
FIG.  395- 

Figs.  396  to  401  represent  the  simplest  forms  of  a  cantilever 
frame.     If   the  member  AB  has  to  support  a  uniformly  dis- 


FIG.  396. 
A        J,         i        4,          -IB 


FIG.  398. 


FIG.  397. 


FIG.  400. 


FIG.   401. 


FIG.  403. 

tributed  load  as  well  as  a  concentrated  load  at  B,  intermediate 
stays  may  be  introduced  as  shown  by  the  full  or  by  the  dotted 


CANTILEVER    TRUSSES.  629 

lines  in  Figs.  398  and  399.  Should  a  live  load  travel  over 
AB,  each  stay  must  be  designed  to  bear  with  safety  the 
maximum  stress  to  which  it  may  be  subjected. 

Figs.  400  and  401  show  cantilever  trusses  with  parallel 
chords.  If  the  truss  is  of  the  double-intersection  type,  Fig. 
401,  the  stresses  in  the  members  terminating  in  B  become  in- 
determinate. They  may  be  made  determinate  by  introducing 
a  short  link  BD,  Fig.  402.  Thus,  if,  in  DB  produced,  BG  be 
taken  to  represent  the  resultant  stress  along  the  link,  and  if 
the  parallelogram  HK  be  completed,  BK  will  represent  the 
stress  along  BE,  and  BH  that  along  BF. 

This  link  device  has  been  employed  to  equalize  the  pressure 
on  the  turn-table  TT  of  a  swing-bridge  (Fig.  403).  An  "  equal- 
izer" or  "  rocker-link"  BD,  Fig.  404,  conveys  the  stresses  trans- 
mitted through  the  members  of  the  truss  terminating  in  D  to 
the  centre  posts  BT. 

Theoretically,  therefore,  the  pressure  over  TT  will  be  evenly 
distributed,  whatever  the  loading  may  be,  if  the  direction  of 
BD  bisects  the  angle  TB T  and  if  friction  is  neglected. 

The  joint  between  the  central  span  and  the  cantilever  re- 
quires the  most  careful  consideration  and  should  fulfil  the 
following  conditions: 

(a)  The  two  cantilevers  should  be  free  to  expand   and  con- 
tract under  changes  of  temperature. 

(b)  The   central  span  should  have  a  longitudinal  support 
which  will  enable  it  to  withstand  the  effect  of  the  braking  of  a 
train  or  the  pressure  of  a  wind  blowing  longitudinally. 

(c)  The    wind-pressure    on    the    central    span    should    bear 
equally  on  the  two  cantilevers. 

(d)  The  connections  at   both   ends  should   have   sufficient 
lateral  rigidity  to  check  undue   lateral  vibration.     Conditions 
(a)  and  (c)  would   be  fulfilled  by   supporting  the  central   span 
like  an  ordinary  bridge-truss  upon  a  rocker  bolted  down  at  one 
end  and  upon   a  rocker   resting  on  expansion   rollers  at  the 
other.     This,  however,  would  not  satisfy  condition  (b).     It  is 
preferable  to  support  the  span  by  means  of  rollers  or  links  at 
both   ends,   and   to   secure   it   to   one   cantilever  only   on    the 
central  line  of  the  bridge  with  a  large  vertical  pin,  adapted  to 


630  THEORY  OF  STRUCTURES. 

transmit  all  the  lateral  shearing  force.  A  similar  pin  at  the 
other  end,  free  to  move  in  an  elongated  hole,  or  some  equiva- 
lent arrangement,  as,  e.g.,  a  sleeve-joint  bearing  laterally  and 
with  rollers  in  the  seat,  is  a  satisfactory  method  of  transmitting 
the  shearing  force  at  that  end  also.  (If  there  is  an  end  post,  it 
may  be  made  to  act  like  a  hinge  so  as  to  allow  for  expansion, 
etc.)  The  points  of  contrary  flexure  of  the  whole  bridge  under 
wind-pressure  are  thus  fixed,  and  all  uncertainty  as  to  wind- 
stresses  removed. 

Where  other  spans  have  to  be  built  adjacent  to  a  large  can- 
tilever span,  it  should  not  be  hastily  assumed  that  it  is  neces- 
sarily best  to  counterbalance  the  cantilever  by  a  contiguous 
cantilever  in  the  opposite  direction.  If  it  is  possible  to  obtain 
good  foundations  arid  if  piers  are  not  expensive,  it  might  be 
cheaper  to  build  a  number  of  short  independent  side  spans  and 
to  secure  the  cantilever  to  an  independent  anchorage.  If  this 
is  done,  care  must  be  taken  to  give  the  abutment  sufficient  sta- 
bility to  take  up  the  unbalanced  thrust  along  the  lower  boom 
of  the  cantilever. 

Suppose  that  the  cantilever  is  anchored  back  by  means  of 
a  single  back-stay. 

Let  W  =  weight  necessary  to  resist  the   pull  of  the  back- 

stay ; 

h  =  depth  of  end  post  of  cantilever  ; 
z  =  horizontal   distance    between    foot   of   post   and 

anchorage  ; 
M—  bending  moment  at  abutment  =  Wz. 

If  it  is  now  assumed  that  the  sectional  areas  of  the  post 
and  back-stay  are  proportioned  to  the  stresses  they  have  to  bear 
(which  is  never  the  case  in  practice),  the  quantity  of  material  in 
these  members  must  be  proportional  to 

!?£*      Wh  =  W*±*  = 


which  is  a  minimum  when  z  — 

If  a  horizontal  member  is  introduced  between  the  feet  of 


CANTILEVER    TRUSSES.  631 

the  back-stay  and  the  post,  the  quantity  of  material  becomes 
proportional  to 

h  h 

which  is  a  minimum  when  z  —  h,  i.e.,  when  the  back-stay  slopes 
at  an  angle  of  45°.  By  making  the  angle  between  the  back- 
stay and  the  horizontal  a  little  less  than  45°,  a  certain  amount 
of  material  may  be  saved  in  the  joints  of  the  back-stays  and 
also  in  the  anchors,  which  more  than  compensates  for  the  in- 
creased weight  of  the  anchors  themselves. 

(Note. — In  these  calculations  it  is  also  assumed  that  the  top 
chord  is  horizontal,  and  that  the  feet  of  the  post  and  back  stay 
are  in  the  same  horizontal  plane.  This  is  rarely  the  case  in 
practice.) 

According  to  the  above  the  weight  of  material  necessary 
for  the  back-stay  is  directly  proportional  to  the  bending  moment 
a±  the  abutment  and  inversely  proportional  to  the  depth  of  the 
cantilever,  other  things  being  equal.  A  double  cantilever  has, 
in  general,  no  anchorage  of  any  great  importance. 

If  the  span  is  very  great,  a  cantilever  bridge  usually  re- 
quires less  material  than  any  other  rigid  structure  of  equal 
strength,  even  though  anchorage  may  have  to  be  provided. 
If  two  large  spans  are  to  be  built,  a.  double  cantilever,  requir- 
ing no  anchorage,  may  effect  a  very  considerable  saving  in 
material,  although  a  double  pier,  of  sufficient  width  for  stability 
under  all  conditions  of  loading,  will  be  necessary. 

Again,  where  false-works  are  costly  or  impossible,  the 
property  of  the  cantilever,  that  it  can  be  made  to  support 
itself  during  erection,  gives  it  an  immense  advantage.  If  the 
design  of  the  cantilever  is  such  that  it  can  be  built  out  rapidly 
and  cheaply,  it  will  often  be  the  most  economical  frame  in  the 
end,  even  if  the  total  quantity  of  material  is  not  so  small  as 
that  required  for  some  other  type  of  bridge.  In  all  engineering' 
work,  quantity  of  material  is  only  one  of  the  elements  of  cost, 
and  this  should  be  carefully  borne  in  mind  when  designing  a 
cantilever  bridge,  because  a  want  of  regard  to  the  method  of 


632  THEORY  OF  STRUCTURES. 

erection  may  easily  add  to  its  cost  an  amount  much  greater 
than  can  be  saved  by  economizing  material. 

In  ordinary  bridge-trusses  the  amount  of  the  web  metal  is 
greatest  at  the  ends  and  least  at  the  centre,  while  the  amount 
of  the  chord  metal  is  least  at  the  ends  and  greatest  at  the 
centre.  Thus,  the  assumption  of  a  uniformly  distributed  dead 
load  for  such  bridges  is,  generally  speaking,  sufficiently  ac- 
curate for  practical  purposes.  In  the  case  of  cantilever 
bridges,  however,  the  circumstances  are  entirely  different.  In 
these  the  amount  of  the  metal  both  in  the  web  and  in  the 
chords  is  greatest  at  the  support  and  least* at  the  end.  For 
example,  the  weight  of  the  cantilevers  (exclusive  of  the  weight 
of  platform,  viz.,  -J  ton  per  lineal  foot)  for  the  Indus  Bridge, 
per  lineal  foot,  varies  from  6J  tons  at  the  supports  to  I  ton  at 
the  outer  ends.  Hence,  the  hypothesis  of  a  uniformly  dis- 
tributed dead  load  for  such  structures  cannot  hold  good. 

The  weight  of  a  cantilever  for  a  given  span  may  be  approxi- 
mately calculated  in  the  following  manner : 

Determine  the  stresses  in  the  several  members,  panel  by 
panel — 

(A)  For  a  load  consisting  of 

(1)  a  given  unit  weight,  say  100  tons,  at  the  outer  end  ; 

(2)  the  corresponding  dead  weight. 

(B)  For  a  load  consisting  of 

(1)  the  specified  live  load  ; 

(2)  the  corresponding  panel  dead  weight. 

Thus,  the  whole  weight  of  a  panel  will  be  the  sum  of  the 
weights  deduced  in  (A)  and  (B),  and  the  total  weight  of  the 
cantilever  will  be  the  sum  of  the  several  panel  weights. 

This  process  evidently  gives  at  the  same  time  the  weights 
of  cantilevers  of  one,  two,  three,  etc.,  panel  lengths,  the  loads 
remaining  the  same. 

The  panel  dead  weights  referred  to  in  (A)  and  (B)  must,  in 
the  first  place,  be  assumed.  This  can  be  done  with  a  large  de- 
gree of  accuracy,  as  the  dead  weight  must  necessarily  gradiially 
increase  towards  the  support,  and  any  error  in  a  particular 
panel  may  be  easily  rectified  by  subsequent  calculations. 


CAN  TILE  VER    TR  USSES. 


633 


Again,  the  preceding  remarks  indicate  a  method  of  finding 
the  most  economical  cantilever  length  in  any  given  case. 

Take,  e.g.,  an  opening  spanned  by  two  equal  cantilevers  and 
an  intermediate  girder.  Having  selected  the  type  of  bridge  to 
be  employed  for  the  intermediate  span,  estimate,  either  from 
existing  bridges  or  otherwise,  the  weights  of  independent 
bridges  of  the  same  type  and  of  different  spans.  Sketch  a 
skeleton  diagram  of  the  cantilever,  extending  over  one-half  of 
the  whole  span,  and  apply  to  it  the  processes  referred  to  in  (A) 
and  (B). 

If  L  is  the  length  of  the  cantilever  and  P  that  of  a  panel, 
the  following  table,  in  which  the  intermediate  span  increases 
by  two  panel  lengths  at  a  time,  may  be  prepared : 


c 
1—  1  1) 

S-,  ±j 

•P-a 

||6 

c 
rt 

u 

rt«2§ 

li:ll 

Jf2.|4 

.11 

0.23 

||| 

'"Id 

llll 

*o  C 

•ss 
|| 

ill? 

JltJl 

Is 

o  c 

3"' 

aT"5 

^  , 

a 

&" 

&"" 

f* 

H°C 

0 

L 

2P 

L-2P 

6P 

L  -  6P 

SP 

L  -  SP 

etc. 

etc. 

Weight  in  col.  3  =  one-half  vi  the  weight  of  the  intermediate 

girder 

+  one-half  of  the  live  load  it  carries  if  uni- 
formly distributed.  (The  proportion  will 
be  greater  than  one-half  for  arbitrarily 
distributed  loads,  and  may  be  easily  de- 
termined in  the  usual  manner.) 
Col.  5  gives  the  weights  obtained  as  in  A. 

weight  on  end  of  cantilever 
Col.  6  =  col.  5  X — -. 

Col.  7  gives  the  weights  obtained  as  in  B. 
Col.  8  =  col.  2  -f-  col.  6  -f  col.  7. 

It  is  important  to  bear  in  mind  that  an  increase  in  the  weight 
of  the  central  span  necessitates  a  corresponding  increase  in  the 


634 


THEORY  OF  STRUCTURES. 


weights  of  the  cantilevers.  Hence,  in  order  that  the  weight  of 
the  structure  may  be  a  minimum,  the  best  material  with  the 
highest  practicable  working  unit  stress  should  be  employed  for 
the  centre  span. 

The  table  must  of  course  be  modified  to  meet  the  require- 
ments of  different  sites.  Thus,  if  anchorage  is  needed,  a  column 
may  be  added  for  the  weights  of  the  back-stays,  etc. 

16.  Curve  of  Cantilever  Boom. — Consider  a  cantilever 
with  one  horizontal  boom  OA,  and  let  x,  y  be  the  co-ordinates 
of  any  point  P  in  the  other  boom,  O  being  the  origin  of  co-or- 

-a J 


\y 


FIG.  405. 


FIG.  406. 


dinates  and  A  the  abutment  end  of  the  cantilever. 

Let  Wbe  the  portion  of  the  weight  of  an  independent  span 
supported  at  O. 

Let  w  be  the  intensity  of  the  load  at  the  vertical  section 
through  P. 

Assume  (i)  that  there  are  no  diagonal  strains,  and,  hence, 
that  the  web  consists  of  vertical  members  only ; 

(2)  that  the  stress  H  in  the   horizontal   boom  is 

constant,  and  therefore  the  bending  moment 
at  P  =  Hy  ; 

(3)  that  the  whole  load  is  transmitted  through  the 

vertical  members  of  the  web. 

Let  k  be  such  a  factor  that  kTl  is  the  weight  of  a  member 
of  length  /,  subjected  to  a  stress  T. 

(Note. — If  /  is  in  feet  and  T  in  tons,  then  k  for  steel  is  about 
.0003,  allowance  being  made  for  loss  of  section  or  increase  of 
weight  at  connections.) 

w  consists  of  two  parts,  viz.,  a  constant  part  p,  due  to  the 
weight  of  the  platform,  wind-bracing,  etc.,  which  is  assumed  to 


CURVE    OF   CANTILEVER  BOOM.  635 

be    uniformly   distributed ;  and    a    variable  part,  due    to    the 
weight  of  the  cantilever,  which  may  be  obtained  as  follows: 
Weight  of  element  dx  of  horizontal  boom  =  kHdx. 

"        "    web  corresponding  to  dx  =  kwydx. 

"        "    element  of  curved  boom  corresponding   to  dx 

-•»£)•«. 

Hence  the  variable  intensity  of  weight 


and 

w  = 


Again,  if  M  is  the  bending  moment  and  5  the  shearing 
force  at  the  vertical  section  through  P,  then 

d'M      ttS  d*y 


kH 


Integrating  twice, 


A  and  B  being  constants  of  integration. 

dv 
When  x  =  o,    y  =  o,     and     ffg-  =  W. 

Thus,  A=o     and  B  =  W. 


636  THEORY   OF  STRUCTURES. 

Hence, 


is  the  equation  to  the  curve  of  the  boom,  and  represents  an 
ellipse  with  its  major  axis  vertical,  and  with  the  lengths  of  the 

(p  +  2kH\* 
two  axes  in  a  ratio  equal  to  ^  —  -777  --  J  . 

The  depth  of  the  longest  cantilever  is  determined  by  the 
vertical  tangent  at  the  end  of  the  minor  axis,  and  corresponds 

to  the  value  of  y  given   by  making  —  =  o  in  the  preceding 

equation,  which  gives  y  =  —. 

For  a  given  value  of  H  the  curve  of  the  boom  is  independ- 
ent of  the  span.  Again,  for  a  given  length  of  cantilever  with 
a  boom  of  this  elliptic  form,  a  value  of  H  may  be  found  which 
will  make  the  total  weight  a  minimum,  and  which  will  there- 
fore give  the  most  economical  depth.  Such  an  investigation, 
however,  can  only  be  of  interest  to  mathematicians,  as  the 
hypotheses  are  far  from  being  even  approximately  true  in 
practice,  and  the  resulting  depth  would  be  obviously  too  great. 

Assumption  (i)  on  page  634  no  longer  holds  when  a  live 
load  has  to  be  considered.  Diagonal  bracings  must  then  be 
introduced,  which  become  heavier  as  the  depth  increases,  in 
consequence  of  their  increased  length.  The  diagonal  bracings 
are  also  largely  affected  by  the  length  of  the  panels.  If  the 
panels  are  short,  and  if  a  great  depth  of  cantilever,  diminishing 
rapidly  away  from  the  abutment,  is  used,  the  angles  of  the 
diagonal  bracing,  near  the  abutment,  will  be  unfavorable  to 
economy.  This  difficulty  may  be  avoided  by  adopting  a 
double  system  of  triangulation  over  the  deeper  part  of  the 
cantilever  only,  or  even  a  treble  system  for  some  distance  in 
a  large  span.  The  objections  justly  urged  against  multiple 
systems  of  triangulation  in  trusses  lose  most  of  their  force  in 
large  cantilevers.  In  the  first  place,  the  method  of  erection 
by  building  out  insures  that  each  diagonal  shall  take  its  proper 
share  of  the  dead  load  ;  and  in  the  second  place,  it  should  be 


CURVE    OF   CANTILEVER   BOOM.  637 

remembered  that  only  in  large  spans  could  a  double  system 
have  anything  to  recommend  it,  and  then  only  near  the  abut- 
ment where  the  stresses  are  greatest :  in  such  cases  the  moving 
load  only  produces  a  small  portion  of  the  entire  stress  in  the 
web.  In  practice,  a  compromise  has  to  be  made  between  dif- 
ferent requirements,  and  the  depth  must  be  kept  within  such 
limits  as  will  admit  of  reasonable  proportions  in  other  respects, 
while  the  diagonal  ties  or  struts  may  be  allowed  to  vary  in  in- 
clination, to  some  extent,  from  one  panel  to  another. 

Again,  in  fixing  the  panel  length,  care  must  be  taken  that 
there  is  no  undue  excess  of  platform  weight,  as  this  will  pro- 
duce ?,  corresponding  increase  in  the  weight  of  the  cantilever. 

An  excessive  depth  of  cantilever  generally  causes  an  in- 
crease in  the  cost  of  erection. 

Both  theory  and  practice,  however,  indicate  that  it  will  be 
more  advantageous  to  choose  a  greater  depth  for  a  cantilever 
than  for  an  ordinary  girder  bridge. 

An  ordinary  proportion  for  a  large  girder  bridge  would  be 
one-ninth  to  one  seventh  of  the  span,  and  if  for  the  girder  were 
substituted  two  cantilevers  meeting  in  the  middle  of  the  span, 
the  depth  might  with  advantage  be  considerably  increased 
beyond  this  proportion  at  the  abutment,  if  it  be  reduced  to  nil 
where  the  cantilevers  meet.  When  a  central  span  is  introduced, 
resting  upon  the  ends  of  the  two  cantilevers,  the  concentrated 
load  on  the  end  gives  an  additional  reason  for  still  further  in- 
creasing the  depth  at  the  abutment  proportionally  to  the  hngth 
of  the  cantilever.  The  greatest  economical  depth  has  probably 
been  reached  in  the  Indus  bridge,  in  which  the  depth  at  the 
abutment  =  .54  X  length  of  cantilever.  Probably  the  propor- 
tion of  one-third  of  the  length  of  the  cantilever  would  be 
ample,  except  where  the  anchorage  causes  a  considerable  part 
of  the  whole  weight,  but  each  case  must  be  considered  on  its 
own  merits.  The  reduction  of  deflection  obtained  by  increas- 
ing the  depth  is  also  an  appreciable  consideration. 

If  a  depth  be  chosen  not  widely  different  from  that  which 
makes  the  quantity  of  material  a  minimum,  the  weight  will  be 
only  slightly  increased,  while  it  is  possible  that  great  structural 
advantages  may  be  gained  in  other  directions.  In  recommend- 


638  THEORY  OF  STRUCTURES. 

ing  a  great  depth  for  a  cantilever  at  its  abutment,  it  is  assumed 
that  the  depth  will  be  continuously  reduced  from  the  abutment 
outwards.  If  the  load  were  continuously  distributed,  it  is  by 
no  means  certain  that  a  cantilever  of  uniform  depth  would  re- 
quire more  material  than  one  of  varying  depth,  but  it  has 
already  been  pointed  out  to  what  extent  the  weight  of  the 
structure  itself  necessarily  varies,  and  if  the  concentrated  load 
at  the  end  were  separately  considered,  the  economical  truss 
would  be  a  simple  triangular  frame  of  very  great  depth.  From 
economic  considerations,  it  would  be  well  to  reduce  the  depth 
of  the  cantilever  at  the  outer  end  to  nil,  but  in  many  cases  it 
is  thought  advisable  to  maintain  a  depth  at  this  point  equal  to 
that  at  the  end  of  the  central  span,  so  that  the  latter  may  be 
built  out  without  false-works,  under  the  same  system  of  erection 
as  is  pursued  in  the  case  of  the  cantilever.  The  post  at  the 
ends  of  the  central  span  and  cantilever  is  sometimes  hinged  to 
allow  for  expansion. 

17.  Deflection. — A  serious  objection  urged  against  can- 
tilever bridges  is  the  excessive  and  irregular  deflection  to  which 
they  are  sometimes  subject.  They  usually  deflect  more  than 
ordinary  truss-bridges,  and  the  deflection  is  proportionately 
increased  under  suddenly  applied  loads.  In  the  endeavor  to 
recover  its  normal  position,  the  cantilever  springs  back  with 
increased  force  and,  owing  to  the  small  resistance  offered  by 
the  weight  and  stiffness  at  the  outer  end,  there  may  result, 
especially  in  light  bridges,  a  kicking  movement.  It  must,  how- 
ever, be  borne  in  mind  that  the  deflection,  of  which  the  impor- 
tance in  connection  with  iron  bridges  has  always  been  recog- 
nized, is  not  in  itself  necessarily  an  evil,  except  in  so  far  as  it  is 
an  indication  or  a  cause  of  over-strain. 

18.  The  Statical  Deflection,  due  to  a  quiescent  load, 
must  be  distinguished  from  what  might  be  called  the  dynamical 
deflection,  i.e.,  the  additional  deflection  due  to  a  load  in  motion. 
The  former  should  not  exceed  the  deflection  corresponding  to 
the  statical  stresses  for  which  the  bridge  is  designed.  The 
amount  of  the  dynamical  deflection  depends  both  upon  the 
nature  of  the  lo^ds  and  upon  the  manner  in  which  they  are 
applied,  nor  are  there  sufficient  data  to  determine  its  value 


LIVE  LOAD.  639 

even  approximately.     It  certainly  largely  increases  the  statical 
stresses  and  produces  other  ill  effects  of  which  little  is  known. 

Hitherto,  the  question  as  to  the  deflection  of  framed  struc- 
tures has  received  but  meagre  attention,  and  formulae  deduced 
for  solid  girders  have  been  employed  with  misleading  results. 
It  would  seem  to  be  more  scientific  and  correct  to  treat  each 
member  separately  and  to  consider  its  individual  deformation. 

19.  Rollers.  —  One  end  of  a  bridge  usually  rests  upon  nests 
of  turned  wrought-iron  or  steel  friction  rollers  running  between 
planed  surfaces.     The  diameter  of  a  roller  should  not  be  less 
than   2   inches,  and   the  pressure  upon  it  in  pounds  per  lineal 
inch   should   not  exceed  500  Vd  if  made  of  wrought-iron,  or 
600  tfa  if  made  of  steel,  d  being  the  diameter  in  inches. 

20.  Live  Load.  —  It  is  a  common  practice  with  many  en- 
gineers to  specify  the  live  load  for  a  bridge  as  consisting  01  a 
number  of  arbitrary  concentrated  weights  which  are  more  or 
less   equivalent  to  the  loads  thrown  upon  the  locomotive  and 
car  axles. 

Figs.  407,  408,  and  409  are  examples  of  such  practice. 


90' 


FIG.  407. 


FIG.  408. 


f 


FIG.  409. 


With  such  a  live  load,  the  determination  of  the  position  of 
the  locomotive  and  cars  which  will  give  a  maximum  shear  and 
a  maximum  bending  moment  at  any  section  is  much  facilitated 
by  the  principles  enunciated  in  Art.  8,  Chap.  II. 


640  THEORY  OF  STRUCTURES. 

If  the  chords*  are  parallel,  and  if  5  is  the  maximum  shear 
transmitted  through  a  diagonal  inclined  at  an  angle  6  to  the 
vertical,  the  maximum  stress  in  that  diagonal  =  vSsec  0,  and 
the  corresponding  stress  transmitted  to  a  chord  through  the 
diagonal 

=  Ssec#sin#  =  5 tan  6. 

A  modification  is  necessary  when  the  chords  are  not  paral- 
lel. Consider,  e.g.,  a  truss  with  a  horizontal  bottom  chord 
and  a  top  chord  composed  of  inclined  members.  Retain  the 
same  notation  as  in  the  article  referred  to,  and  let  />, ,  D9  be 
the  stresses  corresponding  to  \hefirst  and  second  distributions, 
respectively,  in  a  diagonal  met  by  a  vertical  section  between 
the  rth  and  (r  -\-  i)th  weights.  Also,  let  the  member  of  the 
upper  chord  cut  by  the  same  section  be  produced  to  meet  the 
horizontal  chord  produced  in  the  point  C. 


FIG.  410. 

Let  AC  =  h,  and  let  /  be  the  perpendicular  from  C  upon 
the  diagonal  in  question. 
Taking  moments  about  C, 

D,p  —  RJi  —  w, 
and 


—  wr(h  +  /  —  ar  —  x)  —  wr+l(h  +  /  —  ar^  —  x)  —  .  .  . 

-  wr+q(A  +  I  —  ar+q  —  x). 

It.  is  assumed,  for  simplicity,  that  no  weights  leave  or  ad- 
vance upon  the  bridge. 

/.  A      A  , 


LIVE  LOAD. 


641 


according  as 


—  a,)  —  .  .  .  —  wr(k  +  l—ar)  = 


RJt  —  w,(h  +  I  —  a,  —  x)  —  w^(h  +  /  —  a,  —  x)  —  .  .  . 
-  wr(h  -\-l-ar  —  x)  —  tvr+l(h  +  I  —  ar+I  —  x)  —  .  .  . 

-  wr+q(h  +  I  -  ar+q  -f  x\ 


or 


= 


where  Rq'(l  -\-  h)  =  algebraic  sum  of  the  moments,  with  respect 
to   Cf  of  the    weights  transferred, 


and 


Hence, 


= 


according  as 


Take,  e.g.,  the  truss  represented  by  the  accompanying  dia- 
gram (Sault  Ste.  Marie  Bridge),  the  live  load  being  that  shown 
by  Fig.  411,  i.e.,  the  loading  from  a  Standard  Consolidation 
engine  with  four  drivers  and  one  leading  wheel. 


FIG.  411. 


Span  —  239  ft. 

Length  of  centre  verticals  —  40  ft.;  of  end  verticals  —  27  ft, 


THEORY   OF  STRUCTURES. 


Applying  the  principles  referred  to  in  the  preceding  it  is 
found  that  the  distributions  of  live  load,  concentrated  at  the 
panel  points,  which  will  give  the  maximum  stresses  in  the 
several  members,  may  be  tabulated  as  below  : 


End 

Distribu- 

Reac- 

Load 

Load 

Load 

Load 

Load 

Load 

Load 

Load 

Load 

tions. 

tion 

at  A- 

at/2. 

at/8. 

at/>4. 

at  A- 

at/6. 

at/7. 

at/e- 

at/,. 

at  A. 

Case  i  

2  

187990 
162920 

495<» 

38700 
495oo 

45925 
38700 

43750 
45925 

36225 
43750 

36000 
36225 

36000 
36000 

36000 
36000 

36000 
36000 

3  

124230 

6400 

47209 

40200 

43400 

45800 

37100 

36000 

36000 

6400 

47200 

40200 

43400 

45800 

37100 

69410 

6400 

47200 

40200 

4^400 

45800 

37100 

6  

47400 

6400 

47200 

40200 

43400 

45800 

7  

29100 

6400 

47200 

40200 

43400 

"  8  

15380 

6400 

47200 

40200 

Dead  weight 

121500 

27000 

27000 

27000 

27000 

27000 

27000 

27000 

27000 

,27000 

N.B.  —  These  numbers  are  convenient  whole  numbers  within 
about  one-half  of  one  per  cent  of  the  calculated  results.  The 
panel  length  is  also  assumed  to  be  24  ft. 

In  Cases  I  arid  2  the  third  driver  is  at  a  panel  point  ;  in  the 
remaining  cases  the  second  driver  is  at  a  panel  point. 

The  dead  weight  includes  the  weight  of  the  ironwork  and 
flooring.  The  panel  loads  may  be  easily  calculated,  either 
analytically  or  graphically.  For  example,  let  A,  B,  C,  D  be 
four  consecutive  panel  points,  and  let  the  third  driver  be  at  B. 


Panel  load  at  A 


, 
=    7500     g  +  12000 

Panel  load  at  B 


/io8         2\ 

J  =  11823,  say  11,900  Ibs. 


8o+236+288+232 


>  say  49*500  Ibs. 
Panel  load  at  C 


87+21 


288 


=  38445,  say  38,700  Ibs. 


LIVE   LOAD. 


643 


Or,  graphically,  upon  the  vertical  through  B  (Fig.  412)  take 
BM  to  represent  7500  Ibs.,  and  join  AM.  Let  the  vertical 
through  #,  meet  AM  in  £, ,  and  the  horizontal  through  AM 
in  ct.  Then  a^  represents  the  portion  of  7500  Ibs.  borne  at 
B,  and  blcl  the  portion  borne  at  A. 

Also,  take  BN  to  represent  12,000  Ibs.  ;  join  ^4 N,  CN.  Let 
the  verticals  through  a^,  a9,  <z4  meet  AN,  CN  in  b^b^,bi,  and 
the  horizontal  through  ^V  in  £a,  £8,  £4.  Then  #.,&,,  #3^s>  #A 


FIG.  412. 

represent  the  portions  of  each  12,000  Ibs.  borne  at  B,  while 
AA>  ^3  represent  the  portions  borne  at  A,  and  £4£4  the  portion 
borne  at  C. 

Finally,  take  BO  to  represent  10,625  Ibs.,  and  join  CO.  Let 
the  verticals  through  a6,  a6  meet  CO  in  <£6,  ^6,  and  the  horizon- 
tal through  O  in  £6,  c%.  Then  #B^5,  ajb^  are  the  portions  of  each 
10,625  Ibs.  borne  at  B,  while  b^c^,  b6c6  are  the  portions  borne  at 
C.  Thus  the  total  weight  at  B 


=  <*  A  +  "A  +  "A 


A 


A 


It  is  open  to  grave  question  whether  the  extremely  nice 
calculations  required  by  the  assumption  of  arbitrary  weight 
calculations  are  not  unnecessary  except  for  floor  systems.  The 
constantly  increasing  locomotive  and  car  weights  and  the 
variety  in  type  of  locomotive  would  seem  to  render  such  cal- 
culations, based  as  they  are  upon  one  particular  distribution  of 
load,  of  no  effect. 

On  the  other  hand,  if  it  is  assumed  that  the  standard  live 
load  consists  of  a  uniform  load  of,  say,  3000  Ibs.  to  3600  Ibs. 
per  lineal  foot,  with  a  single  weight  of,  say,  25,000  Ibs.  to 
35,000  Ibs.  for  each  truss,  at  the  head  or  at  any  other  specified 


644 


THEORY  OF  STRUCTURES. 


point,  i.e.,  rolling  on  the  uniform  load,  the  calculations  would 
be  much  simplified  and  the  resulting  stresses  would  be  at  least 
as  approximately  accurate.  • 

Let  E  be  the  single  concentrated  load,  T  the  panel  train 
load,  and  D  the  panel  dead  load. 

Consider  a  truss  of  N  panels  with  a  single  diagonal  system, 
Fig.  413,  and  let  E  be  at  the  rth  panel  point. 

N-t     N 


FIG.  413. 

The  shear  immediately  in  front  of  £  due  to  £ 


the  shear  at  same  point  due  to  T 


~  N  2 

the  shear  at  same  point  due  to  D 

D  N(N  —  2r  +  I) 

-W~      ~^     ~* 

Diagonal  Stresses.  —  The  maximum  diagonal  stresses  may 
now  be  easily  tabulated  as  follows  : 

TABLE  T. 


2 

O 

•S-o 

o 

"°            '   J5 

% 

^ 

% 

V 

3 

Q               ^ 

•o 

•o 

6 

•o 

• 

T 

1 

>£ 
flj 

% 

Z 

5 

2  .           | 

«•§           Q 
o 

Diagonal. 

Multiplier  (N  - 

Max.  Vertical  S 
E  transmitted 

(1) 

Multiplier 
(N-r-i 

M 

!p 

Total  Maximut 
Shear  due  to 
transmitted. 

Secant  6. 

Max.  Diagonal  S 
Live  Load. 

Multiplier 
N(N  -  2r 

Vertical  Shear  d 
Load  transmit 
D 
N 
Diag.  Stress  due  t 

j 

fj    t 

A 

N—.N-T 

N-I.N--L   T 

JV    i^ 

N.N—i 

N.N-i  D 

a1 

*w 

2 

2                 N 

1  JV 

2 

2            N 

N-a.N-i  T 

~~l          N 

LIVE  LOAD.  645 

Col.  I  designates  the  several  diagonals. 

Col.  2  gives  the  multiplier  N —  r  for  different  values  of  r. 

Gol.  3  gives  the  maximum  vertical  shears  due  to  E  trans- 
mitted through  the  several  diagonals.  This  shear  for  any 
given  diagonal  is  the  product  of  the  corresponding  multiplier 

in  col.  2  and  -r^r. 

Cols.  4  and  5,  9  and  10  give  similar  quantities  for  the  live 
and  dead  loads. 

Col.  6  gives  the  sums  of  the  shears  in  cols.  3  and  5,  i.e.,  it 
gives  the  total  maximum  vertical  shears  due  to  live  load. 

Col.  8  gives  the  maximum  diagonal  stresses  due  to  live 
load.  For  any  specified  diagonal  it  is  the  product  of  the  cor- 
responding shear  in  col.  7  and  the  secant  of  the  angle  between 
the  vertical  and  the  diagonal  in  question. 

Col.  1 1  in  like  manner  gives  the  maximum  diagonal  stress 
due  to  dead  load. 

Col.  12  gives  the  total  maximum  diagonal  stresses  due  to 
both  live  and  dead  loads. 

Another  column  might  be  added  giving  the  sectional  areas 
of  the  diagonals. 

In  the  above  table  the  diagonal  stresses  due  to  the  live  and 
dead  loads  are  separately  determined,  as  different  coefficients 
of  strength  are  sometimes  specified  for  the  two  kinds  of  load. 
With  a  suitable  compound  coefficient  of  strength,  cols.  6,  8, 
and  ii  may  be  replaced  by  a  column  giving  the  sums  of  the 
corresponding  shears  in  cols.  3,  5,  and  10.  These  sums,  multi- 
plied by  secant  #,  give  the  maximum  diagonal  stresses. 

Stresses  in  the  Verticals. — The  maximum  stress  in  any  ver- 
tical, say  at  the  rth  panel  point,  is  evidently  the  vertical  com- 
ponent of  the  maximum  diagonal  stress  in  the  rth  panel,  i.e.,  it 
is  the  maximum  vertical  shear  in  the  rth  panel. 

To  be  more  accurate,  this  amount  should  be  diminished  by 
the  portion  of  the  weight  of  the  lower  chord  borne  at  the  foot 
of  the  vertical  in  question. 

Chord    Stresses.  —  Take    the    load    at    each    panel    point 

=  ~ 


646 


THEORY  OF  STRUCTURES. 
TABLE   II.     (COMPRESSION  CHORD.) 


- 

T3X! 

„ 

•p 

/• 

N 

C 

1 

I 

+ 

o 

sl+ 

ll*1 

rtC      H 

^ 

- 

11     - 

5*21 

^1 

PH 
I- 

•a 
* 

fl 

•gs^s 

a 
rt 

^6^ 
SsS 

-1 

2 

> 

H 

C/5 

H 

Col.  i  designates  the  chord  panel  length. 
Col.  3  gives  the  several  vertical  shears  transmitted  to  the 
chords    through   the    diagonals.      They   are    the    product    of 


W\W  ~^~  ^  ~^~     i 


Col.  5  gives  the  chord  stresses  due  to  these  shears,  i.e.,  the 
product  of  the  shears  in  col.  3  and  the  corresponding  values  of 
tan  0  in  col.  4. 

Col.  6  gives  the  total  chord  stresses  in  the  several  panels. 
In  any  given  panel  the  total  chord  stress  is  equal  to  the  chord 
stress  due  to  the  shear  in  that  panel  plus  the  total  chord  stress 
in  the  preceding  panel. 

Another  column  for  the  sectional  areas  of  the  several 
lengths  of  chord  may  be  added  if  required,  each  length  being 
designed  as  a  strut,  hinged  or  fixed  at  the  ends,  according  to 
the  method  of  construction. 

A  precisely  similar 
table  may  be  prepared 
for  the  tension  chord. 

EXAMPLE  i.  An 
£7£V^/-panelled  deck-truss 

of  108  ft.  span  and  18  ft.  deep,  with  a  single  diagonal  system. 
Concentrated  load  E  for  each  truss  =  25,000  Ibs. 
Train  load  T  for  each  truss  =  1600  Ibs.  per  lineal  ft. 

=  21,600  Ibs.  per  panel. 

Bridge  (dead)  load  D  for  each  truss  =  800  Ibs.  per  lineal  ft. 

=  10,800  Ibs.  per  panel. 
sec  8  =    ,     tan  6  =  f  . 


c4  a 


LIVE   LOAD.  647 

TABLE   OF   MAXIMUM    DIAGONAL   STRESSES.     (See  TABLE  I.) 


uS 

T 
1 

8 

fx 

•  2 

.2  g-o 

i 

8? 

II 

i 

(I 

co 

II 

u 

« 

"°t3 

i 

00  o 

i 

k 

Q 

k 

8! 

•3?* 

c 

C« 

x^> 

.  *J'd 

•2  *» 
*  rt 

1 

& 

00 

1 

v8    oo 

OWJiJ 

5^3 

0> 

00 

00 

OQ 

Q 

00 

N 

Y 

•1 

r1 

C/) 

S^ 

* 

M 

Q"0 

H 

4 

7 

21875 

21 

56700 

78575 

li 

982,8} 

28 

37800 

4725° 

145468} 

d<i 

6 

18 

7  So 

15 

40500 

5925° 

I* 

74062^ 

20 

27 

000 

33750 

107812$ 

d^ 

5 

is 

62=; 

10 

27000 

42625 

12 

16 

200 

20250 

7353Ii 

<'* 

4 

12 

Soo 

6 

16200 

•28700 

t* 

35875 

4 

5400 

6750 

42625 

1i 

3 

2 

9375 
6250 

3 

i 

8100 
2700 

J7475 
8950 

:j 

21843* 
11187* 

-   4 

—  54°o 
—  16200 

-  6750 
—20250 

15093* 

d, 

' 

3125 

3125 

3906* 

—  20 

—  27000 

-33750 

It  will  be  observed  that  in  the  fifth  panel  there  is  a  maxi- 
mum positive  shear  of  17,475  Ibs.  and  a  negative  shear  of  5400 
IDS.,  the  former  due  to  the  live  and  the  latter  to  the  dead  load. 
The  resultant  shear  of  12,075  IDS-»  which  is  opposite  in  kind  to 
that  due  to  the  dead  load,  is  provided  for  by  means  of  the 
counterbrace  ab.  No  counterbraces  are  theoretically  required 
in  the  sixth  and  seventh  panels,  but  they  are  often  introduced 
in  order  to  stiffen  the  truss. 

TABLE   OF   MAXIMUM   STRESSES   IN  THE   VERTICALS. 

^  =  78575  +  378oo  =  1 16,375  Ibs. 
v^  —  59250  +  27000  =    86,250   " 
i\  —  27000  -|-  16200  =    43,200   " 
v^~  16200+    5400=    21,600   " 

Chord  Stresses. — Load  at  each  panel  point 
=     -+  T+D=  35,525  Ibs. 


TABLE   OF    MAXIMUM    STRESSES    IN   COMPRESSION   CHORD. 


Member. 

4(9  -  ar). 

Sf=<«4 

Tanfl. 

Chord  Stress  due 
to  Shear. 

Total  Max. 
Chord  Stress. 

fl 

28 

I24337i 

1 

93253i 

93253i 

<•* 

2O 

888i2i 

f 

666ogf 

159862^ 

c* 

12 

5328;i 

* 

39965f 

199828^ 

ft 

4 

17762^ 

f 

I332i| 

213150 

648  THEORY   OF  STRUCTURES. 

TABLE   OF   MAXIMUM   STRESSES   IN   TENSION   CHORD. 


Member. 

4(9  -  vr). 

«•-««* 

Tanfl. 

Chord  Stress  due 
to  Shear. 

Total  Max. 
Chord  Stress. 

ti 

28 

124337! 

f 

93253* 

93253* 

ti 

20 

88812! 

f 

666o9f 

159862! 

12 

532871 

f 

39965t 

199828^ 

The  above  figures  may  be  checked  by  the  method  of 
moments. 

Note. — If  the  truss  is  inverted  it  becomes  one  of  the  Howe 
type.  The  stresses  are  the  same  in  magnitude,  but  reversed 
in  kind. 

Ex.  2.  An  eight-panel  through-bridge  of  the  double-inter- 
section type  (Fig.  414),  hav- 
ing the  same  span,  depth, 
and  loading  as  in  Ex.  i. 

The    systems     1234... 
FlG- 4'4-  and    I   a  b  c  .  .  .    are     inde- 

pendent.    It  is  assumed  that  the  load  at  the  foot  of  the  end 
vertical  (gh)  is  divided  equally  between  the  two  systems. 

TABLE   OF   MAXIMUM   STRESSES   IN  END-POSTS   AND 
DIAGONALS. 


•  o 

•o 

m>a 

10 

. 

8 

^  OJ-O 

nf^-d 

1 

L 

il 

v 

m 

_« 

N 

S  3  g 

. 

S  3  § 

fc 

1 

•^ 

c^Q 

si 

a 

II 

71 

II 

u1-^ 

_ 

*^i-3 

'a 

I 

\ 

o  . 

C/5 

a 

V 

H 

O  1 
8  00 

"5 

O  1 
Jloo 

"5J.I 

i 

K^> 

rt's'l 

2.21 

i 

X 

&l 

X 

51 

H^^ 

in 

|c^HJ 

i 

C"0^ 

H5 

! 

7 
6 
5 

4 

21875 
18750 
15625 
12500 

21 

6* 

^} 

56700 
17550 
9450 
6750 

78575 
36300 

25075 
19250 

I 

982I8J 
45375 
45135 
34650 

28 

"i 

4i 

37800 
16875 

"475 
6075 

47250 
21093!- 
20655 
10935 

i45468f 
66465* 
65790 
45585 

*4 

3 

9375 

•J- 

1350 

10725 

1  r 

J93°5 

1. 

)j  C 

1215 

20520 

? 

2 

i 

6250 
1562* 
1562* 

J3S0 

7600 
1562* 
15624 

ii 

13680 
28124 
1953* 

-  34 

~ll* 

-  4725 
—  10125 
-'5525 

-  8505 
-18225 
—  27945 

5'75 

The  counterbrace  cf  is  required  to  take  up  the  resultant 
shear  of  6250  —  4725  =  1525  Ibs.,  which  is  opposite  in  kind  to 
that  due  to  the  dead  load. 

The  first  line  in  the  table  gives  the  maximum  thrust  along 
the  end  post  (/).  It  is  made  up  of  the  stresses  transmitted 


LIVE   LOAD. 


649 


through  the  two  systems  of  diagonals  when  the  25,000  Ibs.  is 
at  the  first  panel  point. 

TABLE   OF   MAXIMUM   STRESSES   IN  VERTICALS. 

The  maximum  stress  in  an  end  vertical  evidently  occurs 
when  the  25,000  Ibs.  is  concentrated  at  its  foot. 

z/j  =  25000  +  10800  =  35800  Ibs.  (tension); 
^—19250+    6075  =  25325    "     (compression); 
v3  =  10725  +      675  =  11400   " 
v4  =    6250—    4725=     1525    "  " 

Chord  Stresses. — Load  at  each  panel  point 
=  ~+T+D  =  35525. 

TABLE  OF  MAXIMUM  STRESSES   IN   COMPRESSION   CHORD. 


Total 

Member 

Multiplier. 

25|£5  =  444of  . 

Tan  0. 

Chord  Stress 
due  to  Shear. 

Maximum 
Chord 

Stress. 

(           28 

I24337& 

§ 

93253* 

j 

Cl 

j      +12* 

55507^1 
37745* 

I 

4163011 
566171* 

|    T9i50i$i 

191501$! 

c? 

4* 

19982^ 

.* 

29974-^ 

221476^5 

Cl 

* 

j 

3330*| 

224806^ 

Note.  —  cl  is  made  up  of  the  thrusts  transmitted  through 


TABLE  OF   MAXIMUM   STRESSES   IN   TENSION   CHORD. 


Member. 

Multi- 
plier. 

35525 

Tang. 

Chord  Stress 
due  to  Shear. 

Total  Maximum 
Chord  Stress. 

8 

/I    =    /S 

28 

124337* 

i 

93253* 

93253^ 

/3 

19} 

i 

'' 

37745T5ir 

i 

56617!* 

191501!* 

Ex.  3.  A  through-bridge   of  the  Warren   type   (Fig.  415) 
having  the  same  span  and  loading  as  in  Exs.  i  and  2. 


FIG.  415. 


650  THEORY  Of  STRUCTURES. 

TABLE   OF    MAXIMUM    STRESSES    IN    DIAGONALS. 


$ 

8 

. 

1 

iii 

1 

0. 

II 

if. 

t^ 

M 
II 

.a 

13. 

II 

-"1 

c. 

11. 

g  0  eg 

i 

u 

1 

8    OO 

"3 

as 

00 

"3 

00     OO 

III 

1 

|II 

d\  ~=-di 

7 

21875 

21 

56700 

28 

37800 

116375 

1.155 

I344I4 

ds—dt 

6 

18750 

15 

40500 

20 

27000 

86250 

99619 

df,  =  d* 

5 

15625 

10 

27OOO 

12 

16200 

58825 

67943 

dT=d* 

4 

12500 

6 

16200 

4 

5400 

34100 

39386 

dg=div 

3 

9375 

3 

8100 

—  4 

-  5400 

12075 

13947 

dn=d^ 

2 

6250 

i 

27OO 

—  12 

—  16200 

d\$~=-d\i 

I 

3125 

—  20 

—27000 

The  resultant  stresses,  d9  =  dl9 ,  are  of  an  opposite  kind  to 
the  corresponding  stresses  due  to  the  dead  load.  Thus,  the 
diagonals  upon  which  they  act  must  be  designed  so  as  to  bear 
both  tensile  and  compressive  stresses.  The  stresses  dlt  ds, 
dt,  .  .  .  are  compressions,  and  d9,  d^ ,  </6,  .  .  .  tensions. 

TABLE    OF    MAXIMUM    STRESSES   IN  COMPRESSION    CHORD. 


Mem- 
ber. 

Multi- 
plier. 

3-f^  =  4440L 

Shear 
transmitted. 

Tanfl. 

Chord  Stress. 

Total 
Maximum 
Chord 
Stress. 

c\ 

j        28 

i      28 

Through  d*  124337! 
d*  124337! 

|  248675 

•577 

143485  + 

•  143486 

Ci 

(        20 

/        2O 

dt  88812! 
d,  88812! 

|  177625 

•577 

102489  + 

245976 

Cz 

<         12 
(         12 

</6  53287! 

^8  53287! 

|  106575 

•577 

61493  + 

307469 

Cl 

\           4 

\        4 

^7  17762! 
<**  17762! 

|    35525 

•577 

20497  + 

32/967 

TABLE   OF   MAXIMUM    STRESSES    IN   TENSION   CHORD. 


Total 

Mem- 
ber. 

Multi- 
plier. 

^f  =  444of 

Shear 
transmitted. 

TanQ. 

Chord  Stress. 

Maximum 
Chord 

Stress. 

/i 

28 

Through  d\  124337! 

124337! 

•577 

71742  + 

7 

/I 

- 

28 
20 

d*  124337! 
d*    88812! 

[213150 

•577 

122987  + 

19. 

*• 

• 

20 
12 

^4    88812! 
d*    53287* 

I  142100 

•577 

81991  + 

276722 

f« 

12 

4 

d*    53287! 
'         di     17762^ 

|    71050 

•577 

40995  + 

3i77i8 

WIND   PRESSURE.  651 

21.  Wind-pressure.  —  Numerous  experiments  to  deter- 
mine the  pressure  and  velocity  of  the  wind  have  been  made 
by  means  of  feathers,  cloud-shadows,  anemometers  of  various 
kinds,  wind-gauges,  pendulum,  tube,  and  spring  instruments. 
The  results,  either  through  errors  of  observation,  errors  of  con- 
struction, or  for  other  occult  reasons,  are  almost  wholly  unre- 
liable and  give  the  engineer  no  accurate  information  upon 
which  to  base  his  calculations  as  to  the  effect  of  wind  upon  a 
structure.  Theoretical  investigations  on  the  subject  are  equally 
unsatisfactory.  The  formulae  expressing  the  relations  between 
the  speed  of  the  anemometer,  the  velocity  of  the  wind  and  its 
pressure,  are  of  a  purely  empirical  character,  and  are  only 
applicable  to  a  specific  series  of  recorded  observations. 

Smeation  inferred  from  Rouse's  experiments  that  the  aver- 
age pressure  in  pounds  per  square  foot  =  (velocity  in  miles  per 
hour)"  -r-  200,  or 
...         :  ,,,..  :  £ 

200 

According  to  Dines  the  formula  should  be 


2000 


The  Wind-Pressure  Commission  (Eng.)  recommended  the 
formula 

F' 
= 


.as  giving  with  tolerable  accuracy  the  maximum  pressure. 

,,Stokes  considers  that  the  actual  wind  velocities  should  be 

^      2.4       4 

•ut  —  =  —  of   the   values   recorded   by   anemometers,    so 

t  a  velocity  of  64  miles  per  hour  recorded  as  corresponding 
t.o:&  maximum  pressure  of  40.6  Ibs.  per  square  foot  (the  aver- 
age oi  five  observed  pressures)  would  be  reduced  to  51.2  miles 
per  hour.  The  average  pressure  corresponding  to  51.2  miles 


652  THEORY  OF  STRUCTURES. 

per  hour  would  be  13.1  Ibs.  per  square  foot  according  to 
Smeaton's  rule  and  only  9.18  Ibs.  according  to  Dines. 

Again,  certain  experiments  at  Greenwich  indicated  that 
the  pressure  was  increased  by  the  stiffness  of  the  copper  wire 
connecting  the  recording  pencil  with  the  pressure  plate,  and  a 
flexible  brass  chain  was  therefore  substituted  for  the  wire. 
Thus  modified,  a  pressure  of  29  Ibs.  per  square  foot  was  regis- 
tered as  corresponding  to  a  velocity  of  64  miles  per  hour, 
whereas  with  the  copper  wire  a  pressure  of  49^-  Ibs.  per  square 
foot  had  been  registered  with  a  velocity  of  only  53  miles  per 
hour. 

These  facts  tend  to  show  that  the  actual  pressure  is  much 
less  than  that  given  by  a  recording  instrument,  and  that  the 
very  high  pressures,  as,  e.g.,  80  Ibs.  per  square  foot  and  even 
more,  must  be  due  to  gusts  or  squalls  having  a  purely  local 
effect.  This  opinion  seems  to  be  confirmed  by  Sir  B.  Baker's 
experiments  at  the  Forth  Bridge,  which  also  indicate  that  the 
pressure  per  square  foot  diminishes  as  the  area  acted  upon 
increases.  No  engineering  structure  could  withstand  a  press- 
ure of  80  Ibs.  per  square  foot  of  surface,  and  a  pressure  of  28 
Ibs.  to  32  Ibs.  would  overturn  carriages,  drive  trains  from  the 
track,  and  stop  all  traffic. 

It  is,  of  course,  well  known  that  wind-forces  sufficiently 
powerful  to  uproot  huge  trees  and  to  demolish  the  strongest 
buildings  are  occasionally  developed  by  whirlwinds,  tornadoes, 
and  cyclones,  but  these  must  be  classed  as  acta  Dei  and  can 
scarcely  be  considered  by  an  engineer  in  his  calculations. 

Numerous  observations  as  to  the  effect  of  wind  upon  struc- 
tures in  different  localities  must  yet  be  made  before  any  useful 
and  reliable  rules  can  be  enunciated.  In  the  case  of  existing 
bridges  the  elongation  of  the  wind-braces  during  a  storm  can 
easily  be  measured  within  -^-^  of  an  inch.  Investigations 
should  be  made  as  to  the  action  of  the  wind  upon  surfaces  of 
different  forms  and  upon  sheltered  surfaces,  as,  e.g.,  upon  the 
surfaces  behind  the  windward  face  in  bridge-trusses.  Again, 
it  is  quite  possible,  if  not  probable,  that  many  of  the  recorded 
upsets  have  been  due  to  a  combined  lifting  and  side  action, 
requiring  a  much  less  flank-pressure  than  would  be  necessary 


EMPIRICAL   REGULATIONS.  653 

if  there  were  no  upward  force,  and  hence  further  light  should 
be  obtained  on  this  point. 

Under  any  circumstances,  the  wind-stresses  should  be  as 
small  as  possible,  compatible  with  safety,  seeing  how  largely 
they  influence  the  sections  of  the  several  members,  especially 
in  bridges  of  long  span. 

22.  Empirical  Regulations. 

Wind-Pressure  Commission  Rules. — For  railway  bridges  and 
viaducts  assume  a  maximum  pressure  of  56  Ibs.  per  square  foot 
upon  an  area  to  be  estimated  as  follows  : 

A.  In  c/ose-girder  bridges  or  viaducts  the  area 

=  area  of  windward  face  of  girder 
-|-  area  of  train  surface  above  the  top  of  the  same  gir- 
der. 

B.  In  0/^-girder  bridges  or  viaducts  the  area  for  the  wind- 

ward girder 
=  area  of  windward  face,  assumed  close,  between  rails 

and  top  of  train 
-|-  calculated  area  of  windward  surface  above  the  top 

of  the  train 

-j-  calculated  area  of  windward  surface  below  the  rails. 
For  the  leeward  girder  or  girders  the  area 

=  calculated  area  of  surface  of  one  girder  above  the 
top  of  the  train  and  below  the  level  of  the  rails, 
the  pressure  being  28,  42,  or  56  Ibs.  per  square 
foot,  according  as  this  area  <  f5,  >  \S  and  <fS, 
or  >{5,  where  5  is  the  total  area  within  the  out- 
line of  the  girder.  The  assumed  factor  of  safety 
is  to  be  4. 

American  Specifications. — (a)  The  lateral  bracing  in  the 
plane  of  the  roadway  is  to  be  designed  so  as  to  bear  a  pressure 
of  30  Ibs.  per  square  foot  upon  the  vertical  surface  of  one 
truss  and  upon  the  surface  of  a  train  averaging  12  sq.  ft.  per 
lineal  foot,  i.e.,  360  Ibs.  per  lineal  foot ;  this  latter  is  to  be  re- 
garded as  a  live  load.  The  lateral  bracing  in  the  plane  of  the 
other  chord  is  to  be  designed  so  as  to  bear  a  pressure  of  50  Ibs. 
per  square  foot  upon  twice  the  vertical  surface  of  one  truss. 
(&)  The  portal,  vertical,  and  horizontal  bracing  is  to  be 


654 


THEORY  OF  STRUCTURES. 


proportioned  for  a  pressure  of  30  Ibs.  per  square  foot  upon 
twice  the  vertical  surface  of  one  truss  and  upon  the  surface  of 
a  train  averaging  10  sq.  ft.  per  lineal  foot,  i.e.,  300  Ibs.  per 
lineal  foot,  the  latter  being  treated  as  a  live  load. 

(c)  Live  load  in  plane    of   roadway  due   to  wind-pressure 

=  300  Ibs.  per  lineal  foot. 
Fixed  load  in  plane  of  roadway  due  to  wind-pressure 

=  150  Ibs.  per  lineal  foot. 
Fixed  load  in  plane  of  other  chord  due  to  wind-pressure 

=  150  Ibs.  per  lineal  foot. 

Lateral  Bracing. — Consider  a  truss-bridge  with  parallel 
chords  and  panels  of  length/.  Let  A  be  the  area  of  the  ver- 
tical surface  of  one  truss. 

According  to  (a),  the  lateral  bracing  in  the  plane  of  the 
roadway  is  subjected  to  (i)  a  panel  live  load  of  360^  Ibs.  and 
(2)  a  panel  fixed  load  of  $oA  Ibs.,  while  in  the  plane  of  the 
other  chord  it  is  subjected  to  a  panel  fixed  load  of 

50  X  2A  —  100^4  Ibs. 

Thus,  if  the  figure  represent  the  bracing  in  the  plane  of  the 
roadway  of  a  ten-panel  truss,  and  if  the  wind  blow  upon  the 


30|A 


30  A 
80JA         36CJjP 


3o|A 

S6£  P 

FlG.  416. 


36g  P       360  P       360  P       36<j[  P 


side  AB,  the  maximum  horizontal  force  for  which  any  diagonalr 
e.g-    CD,  is  to  be  designed  is 

=  4$A  Ibs.  due  to  the  horizontal  force  of  30^  Ibs.  at 

each  panel  point 
+  756^  Ibs.  due  to   the  horizontal   force  of  360^  Ibs.  at 

each  panel  point  between  C  and  B. 

The  dotted  lines  show  the  bracing  required  when  the  wind 
blcws  on  the  opposite  side. 

It   is   sometimes    maintained    that    the  wind-forces  in  the 


CHORDS.  65$ 

plane  of  the  upper  chords  of  a  through-bridge  or  the  lower 
chords  of  a  deck-bridge  are  transmitted  to  the  floor-bracing 
through  the  posts.  This  can  hardly  be  correct  in  the  case  of 
long  posts,  as  they  do  not  possess  sufficient  stiffness.  It  has, 
however,  been  pointed  out  by  Mr.  W.  B.  Dawson  that,  in 
through-bridges,  the  cumulative  effect  of  the  wind-pressure  at 
the  ends  of  the  bridge  might  produce  a  serious  bending  action 
in  the  end  posts.  This  action  would  have  to  be  resisted  by 
additional  plating  on  the  end  posts  below  the  portals,  or  by  an 
increase  of  their  sectional  area. 

Under  wind-pressure  the  floor^beams  act  as  posts  ;  hence, 
if  the  wind-bracing  is  attached  to  the  top  or  compression  flange 
of  a  floor-beam,  the  flange's  sectional  area  must  be  propor- 
tionately increased.  If  the  bracing  is  attached  to  the  lower 
or  tension  flange,  the  stresses  in  the  latter  will  be  diminished. 

23.  Chords.  —  The  wind-pressure  transmitted  through  the 
floor-bracing  increases  the  stresses  in  the  several  members,  or 
panel  lengths,  of  the  leeward  chord,  the  greatest  increments 
being  due  to  a  horizontal  force  of  (360^  -f-  $oA)  Ibs.  at  each  of 
the  panel  points  in  AB.  The  corresponding  chord  stresses  in 
the  ten-panel  truss-bridge  referred  to  above  are  : 

£•,=0; 

C9  =  4^(360;)  -f-  30^)  tan  6  Ibs.  ; 

C3  =  C,  +  3i(3<x>/  +  30^)  tan  8  =  8(360^  +  30^)  tan  V  Ibs.  ; 
C4  —  C3  +  2|(36o/  +  $oA)  tan  B  =  \o%($6op  +  30^)  tan  0  Ibs.  ; 

tan  0  =  \26o  0^    tan  fl  Ibs. 


90°  —  8  being  the  angle  between  a  diagonal  and  a  chord. 

Again,  the  wind-pressure  tends  to  capsize  a  train  and  throws 

y 

an  additional  pressure  of  P-=.  Ibs.  per  lineal  foot  upon  the  lee- 

ward rail,  P  being  the  pressure  in  pounds  per  lineal  foot  on  the 
train  surface,  y  the  vertical  distance  between  the  line  of  action 
of  P  and  the  top  of  the  rails,  and  G  the  gauge  of  the  rails. 


THEORY   OF  STRUCTURES. 
Thus,  the  total  pressure  on  leeward  rail 


( 


w  v\ 

2  +P  G)  lbS>  Pei"  Hneal  f°0t> 


and  the  total  pressure  on  windward  rail 

(w  y\ 

-  —  P  ~j  Ibs.  per  lineal  foot, 

w  being  the  weight  of  the  train  in  pounds  per  lineal  foot. 

Hence,  the  total  vertical  pressure  at  a  panel  point  of  the 
leeward  truss 


S  being  the  distance  between  the  trusses. 

24.  Stringers.  —  Each  length  of  stringer  between  consecu- 
tive floor-beams  may  be  regarded  as  an  independent  girder 
resting  upon  supports  at  the  ends,  and  should  be  designed  to 
bear  with  safety  the  absolute  maximum  bending  moment  to 
which  it  may  be  subjected  by  the  live  load.  If  the  beams  are 
not  too  far  apart,  the  absolute  maximum  bending  moment  will 
be  at  the  centre  when  a  driver  is  at  that  point.  Again,  in  the 
case  of  the  Sault  Ste.  Marie  Bridge,  it  may  be  easily  shown 
that  the  maximum  bending  moment  is  produced  when  the  four 
pairs  of  drivers  are  between  the  floor-beams. 

Let/  —  distance  of  first  driver  from  nearest  point  of  support. 

The  reaction  at  this  support 

=  4tr(824  -  47)  =  4^(206  -y). 

The  bending  moment  is  evidently  a  maximum  at  the  second 
or  third  driver,  and  at  the  second  driver 


—  £o.o.(2o6  —  y)($6  +7)  —  12000  X  56  ; 


MAXIMUM  ALLOWABLE    STRESS. 
at  the  third  driver 

=  104(206  —  7X108  +7)  —  12000(52  +  108). 

In  the  first  case  it  is  an  absolute  maximum  when  y  =  75"  ; 
"     "    second  "     "     "          "  "  "    y  —  49" ; 

its  value  in  each  case  being  2,i88,i66|  in.-lbs. 

Hence,  the  bending  moment  is  an  absolute  maximum  and 
equal  to  2,i88,i66f  in.-lbs.,  at  two  points  distant  75  in.  from 
each  point  of  support. 

Also,  if  /!  is  the  moment  of  inertia  of  the  section  of  the 
stringer  at  these  points,  £,  the  distance  of  the  neutral  axis  from 
the  outside  skin,  and/t  the  coefficient  of  strength,  then 


-(2i88i66f)  —/i  —  for  the  inner  stringer, 


and 


-(2i88i66f)  =  /,-  for  the  outer  stringer. 
3  c\ 

The  continuity  of  the  stringers  adds  considerably  to  their 
strength. 

25.  Maximum  Allowable  Stress.  —  Denoting  by  A  and  B, 
respectively,  the  numerically  greatest  and  least  stresses  to 
which  a  member  is  to  be  subjected,  the  following  rules  will  give 
results  which  are  in  accordance  with  the  best  practice  : 

I.  Members  subjected  to  Tensile  Stresses  only. 

For  ivr  ought-iron,  maximum  stress  per  square  inch 


=  10000  Ibs.  —  8000^1  +  -jj  Ibs.  =  ^3.81  +  1.9  -jj  tons. 

For  steel,  maximum  stress  per  square  inch 
=  12000  Ibs.  =  10000(^1  -f-  -rj  Ibs.  =  [,5.08  -f-  2.54  —  J  tons. 


658  THEORY   OF  STRUCTURES. 

II.  Members  subjected  both  to  Tensile  and  Comprcssivc  Stresses. 

For  wr  ought-iron,  maximum  stress  per  square  inch 

=  8000(1  -  -£jj  Ibs.  =  (3  8l  --  1.9^)  tons. 

For  steel,  maximum  stress  per  square  inch 
=  10000(1  -  £j)  Ibs.  =  (5.08  -  2.54-)  tons. 

III.  Members  subjected  to  Compressive  Stresses  only. 
Denote  the  ratio  of  the  length  (/)  to  the  least  radius  of 

gyration  (k)  by  r. 

The  maximum  stress  per  square  inch  =  —  —  —  2  Ibs., 

/being  8000  Ibs.  for  wrought-iron  and  10,000  Ibs.  for  steel,  and 

—  being  40,000,  30,000,  or  20,000,  according  as  the  member  has 
a 

two  square  (fixed)  ends,  one  square  and  one  pin  end,  or  two 
pin  ends. 

Again,  the  maximum  stress  per  square  inch  for  steel  struts 

(/?  \ 
1  +  A!  ft>s.; 
A  I 


"   "  square  ends  —  (10000—  4Or)  I  +  -j  Ibs.; 

\    A' 

«   "  pin  ends    =  (5  -  ^)(i  +  ~)  tons  ; 


"      "     square  ends  =  (  5  —  ^Jl1  4~  ~/r]  tons. 

In  the  last  two  expressions  r  <  40.     These  expressions  may 
be  also  employed  in  the  case  of  alternating  stresses,  but  the 

(T) 
I    -| 


CAMBER.  659 

26.  Camber.  —  Owing  to  the  play  at  the  joints,  a  girder  or 
truss  will  deflect  to  a  much  greater  extent  than  is  indicated  by 
theory,  and  the  material  will  receive  a  permanent  set,  which, 
however,  will  not  prove  detrimental  to  the  stability  of  the 
structure  unless  it  is  increased  by  subsequent  loads.  If  the 
chords  were  initially  made  straight,  they  would  curve  down- 
wards ;  and  although  it  does  not  necessarily  follow  that  the 
strength  of  the  truss  would  be  sensibly  impaired,  the  appear- 
ance would  not  be  pleasing. 

In  practice  it  is  often  specified  that  the  girder  or  truss  is  to 
have  such  a  camber  or  upward  convexity  that  under  ordinary 
loads  the  grade  line  will  be  true  and  straight  ;  or,  again,  that  a 
camber  shall  be  given  to  the  span  by  making  the  panel  lengths 
of  the  top  chord  greater  than  those  of  the  bottom  chord  by 
.125  in.  for  every  10  ft. 

The  lengths  of  the  web  members  in  a  cambered  truss  are 
not  the  same  as  if  the  chords  were  horizontal,  and  must  be  care- 
fully calculated  so  as  to  insure  that  the  several  parts  will  fit 
together. 

To  find  an  Approximate  Value  for  the  Camber,  etc. 

Let  d  be  the  depth  of  the  truss. 

Let  s1  ,  sy  be  the  lengths  of  the  upper  and  lower  chords,  re- 
spectively. 

Let  /"j  ,  yj  be  the  unit  stresses  in  upper  and  lower  chords, 
respectively. 

Let  dl  ,  d^  be  the  distances  of  the  neutral  axis  from  the 
upper  and  lower  chords,  respectively. 

Let  R  be  the  radius  of  curvature  of  the  neutral  axis. 

Let  /  be  the  span  of  the  truss, 

Then 

4  £-/  /I  4  /-*.  /, 

=  -  =      and     =  -   = 


the  chords  being  assumed  to  be  circular  arcs. 

Hence,  the  excess  in  length  of  the  upper  over  the  lower 
chord 


660  THEORY  OF  STRUCTURES. 

Let  ;r, ,  x^  be  the  cambers  of  the  upper  and  lower  chords, 
respectively ;  R  -f-  d^  and  R  —  d^  are  the  radii  of  the  upper 
and  lower  chords,  respectively. 

By  similar  triangles, 

the  horizontal  distance  between  )        R-\-  d 


the  ends  of  the  upper  chord      j   "          R 

I  ._  K-a 
\-—^L 


the  horizontal  distance  between  i        R  — 
the  ends  of  the  lower  chord 


Hence, 

\1     " 
and 


j  =  x,  .  2(R  +  d,\  approximately, 
(-  —  ^  —  -/)  =  x^  .  2(R  —  d^),  approximately. 


, 
and    **  =  SRI--R]- 

27.    Rivet-connection   between    Flanges   and  Web.— 

The  web  is  generally  riveted  to  angle-irons  forming  part  of 
the  flanges. 

The  increment  of  the  flange  stress  transmitted  through 
the  web  from  point  to  point  tends  to  make  the  angle-irons  slide 
over  the  flange  surfaces. 

Denote  the  increment  by  F,  and  let  h  be  the  effective  depth 
of  the  girder  or  truss. 

Then,  if  5  be  the  shearing  force  at  any  point, 

Fh  =  the  increment  of  the  bending  moment  per  unit 
of  length 

ldM\ 
=  l—  =  —    =  o  in  the  case  ot  a  close  web, 

\dx  I 

and        Fh  —  the  increment  of  the  bending  moment 

=  (4M)  =  Sa  in  the  case  of  an  open  web  ; 

a  being  the  distance  between  the  two  consecutive  apices  or 
panel  points  within  which  ^  lies. 


EYE-BARS  AND   PINS. 


66 1 


Hence,  if  TV  be  the  number  of  rivets  per  unit  of  length  for 
the  close  web,  or  the  number  between  the  two  consecutive 
apices  for  the  open  web, 

N—  —fs  =  F  =  Y  for  the  close  web, 


and 


=  —T-  for  the  open  web, 


d  being  the  diameter,  of  a  rivet,  and  fs  the  safe  coefficient  of 
shearing  strength. 

28.  Eye-bars  and  Pins. — Eye-bars  connected  with  pins 
have  been  commonly  employed  in  the  construction  of  suspen- 
sion cables,  the  tension  chords  of  ordinary  trusses  and  canti- 
levers, and  the  diagonals  of  web  systems.  The  requisite  sec- 
tional area  is  obtained  by  placing  a  number  of  bars  side  by 
side  on  the  same  pin,  and,  if  necessary,  by  setting  two  or  more 
tiers  of  bars  one  above  another. 


FIG.  417. 


rFh 


rf=h 


FIG.  418. 


FIG.  419. 


The  figures  represent  groups  of  eye-bars  as  they  often 
occur  in  practice. 

If  two  sets  of  2n  bars  pull  upon  the  pin  in  opposite  direc- 
tions, as  in  Figs,  418  and  419,  the  bending  moment  on  the  pin 
will  be  nPp,  P  being  the  pull  upon  each  bar,  and  /  the  distance 
between  the  centre  lines  of  two  consecutive  bars. 


THEORY  OF  STRUCTURES. 


Hence, 


f  being  the  stress  in  the  material  of  the  pin  at  a  distance  c  from 
the  neutral  axis,  and  /  the  moment  of  inertia. 

In  general,  the  bending  action  upon  a  pin  connecting  a 
number  of  vertical,  horizontal,  and  inclined  bars  may  be  de- 
termined as  follows  : 

Consider  one-half  of  the  pin  only. 

Let  F,  Fig.  420^  be  the  resultant  stress  in  the  vertical  bars. 
It  is  necessarily  equal  in  magnitude  but  opposite  in  direc- 
tion to  the  vertical  component  of  the  resultant  of  the  stresses 


h-*- 


FIG.  420. 

in  the  inclined  bars.  Let  v  be  the  distance  between  the  lines 
of  action  of  these  two  resultants.  The  corresponding  bending 
action  upon  the  pin  is  that  due  to  a  couple  of  which  the  mo- 
ment is  Vv. 

Let  h  be  the  distance  between  the  lines  of  action  of  the 
equal  resultants  H  of  the  horizontal  stresses  upon  each  side  of 
the  pin.  The  corresponding  bending  action  upon  the  pin  is 
that  due  to  a  couple  of  which  the  moment  is  Hh. 

Hence,  the  maximum  bending  action  is  that  due  to  a  couple 
of  which  the  moment  is  the  resultant  of  the  two  momenjts  Vv 
and  Hh.  viz., 


Eye-bars. — In  England  it  has  been  the  practice  to  roll  bars 
having  enlarged  ends,  and  to  forge  the  eyes  under  hydraulic 


EYE-BARS  AND   PINS.  663 

pressure  with  suitably  shaped  dies.  In  America  both  hammer- 
forged  and  hydraulic-forged  eye-bars  are  made,  the  latter  being 
called  weldless  eye-bars.  Careful  mathematical  and  experimen- 
tal investigations  have  been  carried  out  to  determine  the  proper 
dimensions  of  the  link-head  and  pin,  but  owing  to  the  very 
complex  character  of  the  stresses  developed  in  the  metal  around 
the  eye,  an  accurate  mathematical  solution  is  impossible. 

Let  d  be  the  width  and  / 
the  thickness  of  the  shank  of 
the  eye-bar  represented  in  Fig. 
421.  Let  5  be  the  width  of 
the  metal  at  the  sides  of  the 
eye,  and  H  the  width  at  the 
end.  Let  D  be  the  diameter 
of  the  pin. 

The  proportions  of  the  head 
are  governed  by  the  general  condition  that  each  and  every  part 
should  be  at  least  as  strong  as  the  shank. 

When  the  bar  is  subjected  to  a  tensile  stress  the  pin  is 
tightly  embraced,  and  failure  may  arise  from  any  one  -of  the 
following  causes  : 

(a)   The  pin  may  be  shorn  through. 

Hence,  if  the  pin  is  in  double  shear,  its  sectional  area  should 
be  at  least  one-half  that  of  the  shank. 

It  may  happen  that  the  pin  is  bent,  but  that  fracture  is  pre- 
vented by  the  closing  up  of  the  pieces  between  the  pin-head 
and  nut ;  the  efficiency,  however,of  the  connection  is  destroyed, 
as  the  bars  are  no  longer  free  to  turn  on  the  pin. 

In  practice,  D  for  flat  bars  varies  from  f^to  %d,  but  usually 
lies  between  %d  and  ^d. 

The  diameter  of  the  pin  for  the  end  of  a  round  bar  is  gen- 
erally made  equal  to  \\  times  the  diameter  of  the  bar. 

The  pin  should  be  turned  so  as  to  fit  the  eye  accurately, 
but  the  best  practice  allows  a  difference  of  from  -fa  to  ^4^  °f 
an  inch  in  the  diameters  of  the  pin  and  eye. 

(&)   The  link  may  tear  across  MN. 

On  account  of  the  perforation  of  the  head,  the  direct  pull 
on  the  shank  is  bent  out  of  the  straight  and  distributed  aver 


664  THEORY  OF  STRUCTURES. 

the  sections  S.  There  is  no  reason  for  the  assumption  that 
the  distribution  is  uniform,  and  it  is  obviously  probable  that 
the  intensity  of  stress  is  greatest  in  the  metal  next  the  hole. 
Hence,  the  sectional  area  of  the  metal  across  MN  must  be  at 
least  equal  to  that  of  the  shank,  and  in  practice  is  always 
greater. 

5  usually  varies  from  .55^  to  .625^. 

The  sectional  area  through  the  sides  of  the  eye  in  the  head  of 
a  round  bar  varies  from  i-J  times  to  twice  that  of  the  bar. 

(c)  The  pin  may  be  torn  through  the  head. 

Theoretically,  the  sectional  area  of  the  metal  across  PQ 
should  be  one-half  that  of  the  shank.  The  metal  in  front  of 
the  pin,  however,  may  be  likened  to  a  uniformly  loaded  girder 
with  both  ends  fixed,  and  is  subjected  to  a  bending  as  well  as 
to  a  shearing  action.  Hence,  the  minimum  value  of  H  has  been 
fixed  at  \d,  and  if  H  is  made  equal  to  d,  both  kinds  of  action 
will  be  amply  provided  for. 

(d)  The  bearing  surface  may  be  insufficient. 

If  such  be  the  case,  the  intensity  of  the  pressure  upon  the 
bearing  surface  is  excessive,  the  eye  becomes  oval,  the  metal  is 
upset,  and  a  fracture  takes  place.  Or  again,  as  the  hole  elon- 
gates, the  metal  in  the  sections  5  next  the  hole  will  be  drawn 
out,  and  a  crack  will  commence,  extending  outwards  until  frac- 
ture is  produced. 

In  practice,  adequate  bearing  surface  may  be  obtained  by 
thickening  the  head  so  as  to  confine  the  maximum  intensity  of 
the  pressure  within  a  given  limit. 

(e)  The  head  may  be  torn  through  the  shoulder  at  XY. 
Hence,  ^TFis  made  equal  to  d. 

The  radius  of  curvature  R  of  the  shoulder  varies  from  I \d 
to  J.6d. 

Note. — The  thickness  of  the  shank  should  be  — ,  or  —d  at 

4          7 
least. 

The  following  table  gives  the  eye-bar  proportions  common 
in  American  practice  : 


STEEL   EYE-BARS. 


665 


Value  of  6". 

of  d. 

of  D. 

Weldless 

Hammered 

Bars. 

Bars. 

.00 

•67 

1-5 

•33 

.00 

•75 

1-5 

•33 

.00 

1.  00 

1-5 

•50 

.00 

1.25 

1.6 

•50 

.00 

1-33 

1-7 

.00 

1.50 

1.85 

.67 

.00 

i-75 

2.OO 

•67 

.00 

2.OO 

2.25 

•75 

Also,  in  weldless  bars,  //=  S\  in  hammered  bars,  H '  =  d. 

29.  Steel  Eye-bars. — Hydraulic-forged  steel  eye-bars  are 
now  being  largely  made.  The  steel  has  an  ultimate  tenacity  of 
from  60,000  to  68,000  Ibs.  per  square  inch,  an  elastic  limit  of  not 
less  than  50  per  cent,  and  an  elongation  of  from  17  to  20  per  cent 
in  a  length  equal  to  ten  times  the  least  transverse  dimension. 

The  Phcenix  Bridge  Co.  and  the  Edge  Moor  Iron  Co.  give 
the  following  tables  of  steel  eye-bar  proportions : 


Phoenix  Bridge  Co. 

Edge  Moor  Iron  Co. 

Mini-     i      Excess  of 

Width  of 
bar*/. 

Diameter  of 
Pin-hole. 

Diam- 
eter of 
Head. 

Width 
of 
baroT. 

Diameter 
of 
Pin-hole. 

Diameter 
of 
Head. 

mum 
Thick- 
ness 

Sectional  Area 
of  Head  along 
PP  over  Sec- 

of  Bar.   1    tion  of  Bar. 

3 

2&.  2lf 

7 

2 

Ii 

*4 

I 

33# 

3 

3iV  3lf 

8 

2 

2| 

si 

I 

33* 

4 

3TV 

9 

2i 

2i 

5+ 

I 

33# 

4 

311,  4rV  4tt 

10 

2i 

3i 

6i             i 

33# 

5 
5 
5 

3tf  '  4¥ 
4H«  5TV 

sl-5,  V* 

ii 

12 
13 

3 

3 
4 

2i 

4 
4f 

6*             £ 
8               f 
9*             f 

33# 
33# 
33£ 

6 

4H-  4lt 

13 

4 

3 

ioi             f 

33^ 

6 

5f«,  5iV  5lf 

14 

5 

41 

«i             f 

37^ 

6 

6ft,  6tt.  6Tf 

15 

5 

5| 

I2i 

f 

37^ 

7 

5  A 

15 

6 

5i 

I3i 

| 

37£ 

7 

5lf,  6ft.  6H 

16 

6 

6i 

J4i 

37^ 

8 

6'if,  7ft,  7U 
6f* 

17 
17 

7 

7 

5* 

7i 

1  51 
17 

i 

40^ 
40^ 

8 

6ft,  6tf,  7A 

18 

8 

5f 

17 

i 

40^ 

8 

7li  8f 

19 

8 

6f 

18 

i 

4<$ 

8 

ai,9t 

20 

9 
9 

7T\-  7H 
8|,  8| 

2O 
21 

10 

8| 

22 

10 

8J,9f 

23 

10 

10,    10^ 

24 

In  both  the  Phoenix  and  Edge  Moor  bars  the  thickness  of  the  head  is  the 
same  as  that  of  the  body  of  the  bar,  or  does  not  exceed  it  by  more  than  -fa  in. 


O66  THEORY  OF  STRUCTURES. 

30.  Rivets. — A  rivet  is  an  iron  or  steel  shank,  slightly 
tapered  at  one  end  (the  tail),  and  surmounted  at  the  other  by 
a  cup  or  pan-shaped  head  (Fig.  422).  It  is  used  to  join  steel  or 
iron  plates,  bars,  etc.  For  this  purpose  the  rivet  is  generally 
heated  to  a  cherry-red,  the  shank  or  spindle  is  passed  through 


D 


FIG.  422.  FIG.  423.  FIG.  424.  FIG.  425.  FIG.  426. 

the  hole  prepared  for  it,  and  the  tail  is  made  into  a  button,  or 
point.  The  hollow  cup-tool  gives  to  the  point  a  nearly  hemi- 
spherical shape,  and  forms  what  is  called  a  snap-rivet  (Fig. 
423).  Snap-rivets,  partly  for  the  sake  of  appearance,  are  com- 
monly used  in  girder-work,  but  they  are  not  so  tight  as  conical- 
pointed  rivets  (staff-rivets),  which  are  hammered  into  shape 
until  almost  cold  (Fig.  424). 

When  a  smooth  surface  is  required,  the  rivets  are  counter-sunk 
(Fig.  425).  The  counter-sinking  is  drilled  and  may  extend 
through  the  plate,  or  a  shoulder  may  be  left  at  the  inner  edge. 

Cold-riveting  is  adopted  for  the  small  rivets  in  boiler  work 
and  also  wherever  heating  is  impracticable,  but  tightly-driven 
turned  bolts  are  sometimes  substituted  for  the  rivets.  In  all 
such  cases  the  material  of  the  rivets  or  bolts  should  be  of  su- 
perior quality. 

Loose  rivets  are  easily  discovered  by  tapping,  and,  if  very 
loose,  should  be  at  once  replaced.  It  must  be  borne  in  mind, 
however,  that  expansions  and  contractions  of  a  complicated 
character  invariably  accompany  hot-riveting,  and  it  cannot  be 
supposed  that  the  rivets  will  be  perfectly  tight.  Indeed,  it  is 
doubtful  whether  a  rivet  has  any  hold  in  a  straight  drilled  hole, 
except  at  the  ends. 

Riveting  is  accomplished  either  by  hand  or  machine,  the  latter 
being  far  the  more  effective.  A  machine  will  squeeze  a  rivet, 
at  almost  any  temperature,  into  a  most  irregular  hole,  but  the 
exigencies  of  practical  conditions  often  prevent  its  use,  except 
for  ordinary  work,  and  its  advantages  can  rarely  be  obtained 


STRENGTH  OF  PUNCHED  AND  DRILLED  PLATES.     667 

where  they  would  be  most  appreciated,  as,  e.g.,  in  the  riveting 
up  of  connections. 

31.  Dimensions  of  Rivets.  —  The  diameter  (d)  of  a  rivet 
in  ordinary  girder-work  varies  from  f  in.  to  I  inch,  and  rarely 
exceeds  i£  in. 

The  thickness  (/)  of  a  plate  in  ordinary  girder-work  should 
never  be  less  than  J  in.,  and  a  thickness  of  f  in.,  or  even  -^  in., 
is  preferable. 

Let  T  be  the  total  thickness  through  which  a  rivet  passes. 
According  to  Fairbairn, 

When  /  <  %  in.,  d  should  be  about  2t. 
When  /  >  -J  in.,  d  should  be  about  \\t. 
According  to  Unwin, 

When  /  varies  from  J  in.   to    I    in.   and   passes  through 
two  thicknesses  of  plate,  d  lies  between  f  t  -f-  jV  and 


When  the  rivets  join  several  plates,  d  —  --  (-  —  . 

8       8 

According  to  French  practice, 
Diameter  of  head  —  if  d. 
Length  of  rivet  from  head  =  T  -f-  \\d. 

According  to  Rankine, 

Length  of  rivet  from  head  =  T  -f-  2\d. 
The  rise  of  the  head  =  \d. 

The-  diameter  of  the  rivet-hole  is  made  larger  than  that  of 
the  shank  by  from  ^  to  -J  in.,  so  as  to  allow  for  the  expansion 
of  the  latter  when  hot. 

There  seems  to  be  no  objection  to  the  use  of  long  rivets, 
provided  they  are  properly  heated  and  secured. 

32.  Strength  of  Punched  and  Drilled  Plates.  —  Experi- 
ment shows  that  the  tenacity  of  iron  and  steel  plates  is  con- 
siderably diminished  by  punching.  This  deterioration  in 
tenacity  seems  to  be  due  to  a  molecular  change  in  a  narrow 
annulus  of  the  metal  around  the  hole.  The  removal  of  the 
annulus  largely  neutralizes  the  effect  of  the  punching,  and, 
hence,  the  holes  are  sometimes  punched  -J  in.  less  in  diameter 
than  the  rivets  and  are  subsequently  rimered  or  drilled  out  to 
the  full  size.  The  original  strength  may  also  be  almost 


668  THEORY   OF   STRUCTURES. 

entirely  restored  by  annealing,  and,  generally,  in  steel  work, 
either  this  process  is  adopted  or  the  annulus  referred  to  above 
is  removed. 

Punching  does  not  sensibly  affect  the  strength  of  Landore- 
Siemens  unannealed  plates,  and  only  slightly  diminishes  the 
strength  of  thin  steel  plates,  but  causes  a  considerable  loss  of 
tenacity  in  thick  steel  plates  ;  the  loss,  however,  is  less  than 
for  iron  plates. 

The  harder  the  material  the  greater  is  the  loss  of  tenacity. 

Iron  seems  to  suffer  more  from  punching  when  the  holes 
are  near  the  edge  than  when  removed  to  some  distance  from 
it,  while  mild  steel  suffers  less  when  the  hole  is  one  diameter 
from  the  edge  than  when  it  is  so  far  that  there  is  no  bulging  at 
the  edge. 

The  injury  caused  by  punching  may  be  avoided  by  drilling 
the  holes.  In  important  girder-work  and  whenever  great 
accuracy  of  workmanship  is  required,  a  uniform  pitch  may  be 
insured  and  the  full  strength  of  the  metal  retained  by  the  use 
of  multiple  drills.  Drilling  is  a  necessity  for  first-class  work 
when  the  diameter  of  the  holes  is  less  than  the  thickness  of 
the  plate,  and  also  when  several  plates  are  piled.  It  is  impos- 
sible to  punch  plates,  bars,  angles,  etc.,  in  spite  of  all  ex- 
pedients, in  such  a  manner  that  the  holes  in  any  two  exactly 
correspond,  and  the  irregularity  becomes  intensified  in  a  pile, 
the  passage  of  the  rivet  often  being  completely  blocked.  A 
drift,  or  rimer,  is  then  driven  through  the  hole  by  main  force, 
cracking  and  bending  the  plates  in  its  passage,  and  separating 
them  one  from  another. 

The  holes  may  be  punched  for  ordinary  work,  and  in  plates 
sf  which  the  thickness  is  less  than  the  diameter  of  the  rivets. 
\Vhenever  the  metal  is  of  an  inferior  quality,  the  holes  should 
be  drilled. 

33.  Riveted  Joints. — In  lap  joints  (Figs.  427  and  430)  the 
plates  overlap  and  are  riveted  together  by  one  or  more  rows 
of  rivets  which  are  said  to  be  in  single  shear,  as  each  rivet  has 
to  be  sheared  through  one  section  only. 

In  fish  (or  btitf)  joints  (Figs.  428  and  429)  the  rivets  are 
in  double  shear,  i.e.,  must  be  each  sheared  through  two  sections. 


RIVETED  JOINTS. 


669 


Thus  they  are   not  subjected  to  the  one-sided  pull  to  which 
rivets  in  single  shear  are  liable. 


FIG.  427 


In  fish  joints  the  ends  of  the  plates  meet,  and  the  plates 
are  riveted  to  a  single  cover  (Fig.  428),  or  to  two  covers  (Fig. 
429),  by  means  of  one  or  more  rows  of  rivets  on  each  side  of  the 
joint. 

A  fish  joint  is  properly  termed  a  butt  joint  when  the  plates 
are  in  compression.  The  plates  should  butt  evenly  against  one 
another,  although  they  seldom  do  so  in  practice.  Indeed,  the 
mere  process  of  riveting  draws  the  plates  slightly  apart,  leav- 
ing a  gap  which  is  often  concealed  by  caulking.  A  much 
better  method  is  to  fill  up  the  space  with  some  such  hard  sub- 
stance as  cast-zinc,  but  the  best  method,  if  the  work  will  allow 
of  the  increased  cost,  is  to  form  a  jump  joint,  i.e.,  to  plane  the 
eyes  of  the  plates  carefully,  and  then  bring  them  into  close 
contact,  when  a  short  cover  with  one  or  two  rows  of  rivets  will 
suffice  to  hold  them  in  position. 

The  riveting  is  said  to  be  single,  double,  triple,  etc.,  according 
as  the  joint  is  secured  by  one,  two,  three,  or  more  rows  of  rivets. 


o  o  o 
o  o  o 

O   0   O 

o  o  o 
o  o  o 
o  o  o 

°^o 

0  0 

0°0 

0°0 

0°0 

o°o 

ZIGZAG 

FIG.  432. 

CHAIN 
FIG.  431. 


Double,  triple,  etc.,  riveting  may  be  chain  (Fig.  431)  or  zig- 
zag (Fig.  432).  In  the  former  case  the  rivets  form  straight 
lines  longitudinally  and  transversely,  while  in  the  latter  the 
rivets  in  each  row  divide  the  space  between  the  rivets  in  adja- 
cent rows.  Experiments  indicate  that  chain  is  somewhat 
stronger  than  zigzag  riveting. 


670 


THEORY   OF  STRUCTURES. 


Figs.  433  to  435  show  forms  of  joint  usually  adopted  for 
bridge-work.  In  boiler-work  the  rivets  are  necessarily  very 
close  together,  and  if  the  strength  of  the  solid  plate  be  assumed 
to  be  IOO,  the  strength  of  a  single-riveted  joint  hardly  exceeds 
50,  while  double-riveting  will  only  increase  it  to  60  or  70.  Fair- 


o  o 

o  o  o 

o  o  o 

o  o 


FIG.  433. 


FIG.  434. 


FIG. 


435- 


bairn  proposed  to  make  the  joint  and  unpunched  plate  equally 
strong  by  increasing  the  thickness  of  the  punched  portion  of 
the  plate,  but  this  is  somewhat  difficult  in  practice. 

The  stresses  developed  in  a  riveted  joint  are  of  a  most  com- 
plex character  and  can  hardly  be  subjected  to  exact  mathe- 
matical analysis.  For  example,  the  distribution  of  stress  will 
be  necessarily  irregular  (a)  if  the  pull  upon  the  joint  is  one- 
sided ;  (b)  when  local  action  exists,  or  the  plates  stretch,  or  in- 
ternal strains  are  in  the  metal  before  punching ;  (c)  if  there  is  a 
lack  of  symmetry  in  the  arrangement  of  the  rivets,  so  that  one 
rivet  is  more  severely  strained  than  another;  (d)  when  the 
workmanship  is  defective. 

The  joint  may  fail  in  any  one  of  the  following  ways  : 

(1)  The  rivets  may  shear. 

(2)  The  rivets  may  be  forced  into  and  crush  the  plate. 

(3)  The  rivets  may  be  torn  out  of  the  plate. 

(4)  The  plate  may  tear  in  a  direction  transverse  to  that  of 
the  stress. 

The  resistance  to  rupture  should  be  the  same  in  each  of  the 
four  cases,  and  always  as  great  as  possible. 

The  shearing  and  tensile  strengths  of  plate-iron  are  very 
nearly  equal.  Thus,  iron  with  a  tenacity  of  20  tons  per  square 
inch  has  a  shearing  strength  of  18  to  20  tons  per  square  inch. 
Rivet-iron  is  usually  somewhat  stronger  than  plate-iron. 

Again,  the  shearing  strength  of  steel  per  square  inch  varies 


THE  ORE  TIC  A  L  DED  UC  TIONS.  67 1 

from  about  24  tons  for  steel,  with  a  tenacity  of  about  30  tons, 
to  about  33  tons  for  steel,  with  a  tenacity  of  about  50  tons ;  an 
average  value  for  rivet-steel  with  a  tenacity  of  30  tons  being 
24  tons. 

Hence,  if  4  be  a  factor  of  safety,  the  working  coefficient? 
become 

For  wrought-iron  \  *  tons  Per  SC*uare  inch  in  shear'  and 
(  5     "       "  "      ';  tension. 

_             .  (6  tons  per  square  inch  in  shear,  and 

F°rSteel \n«       "  "      "tension 

Allowance,  however,  must  be  made  for  irregularity  in  the  dis- 
tribution of  stress  and  for  defective  workmanship,  and  in 
riveting  wrought-iron  plates  together  it  is  a  common  practice 
to  make  the  aggregate  section  of  the  rivets  at  least  equal  to 
and  sometimes  20  per  cent  greater  than  the  net  section  ©f  the 
plate  through  the  rivet-holes. 

Hence,  the  working  coefficients  are  reduced  to 

4  or  4^  tons  per  square  inch  for  wrought-iron, 
and 

5  or  5J     "       "         "          "      "    steel, 

according  to  the  character  of  the  joint. 

There  is  very  little  reliable  information  respecting  the  in- 
dentation of  plates  by  rivets  and  bolts,  and  it  is  most  uncertain 
to  what  extent  the  tenacity  of  the  plates  is  affected  by  such 
indentation.  Further  experiments  are  required  to  show  the 
effect  of  the  crushing  pressure  upon  the  bearing  area  (i.e.,  the 
diameter  of  the  rivet  multiplied  by  the  thickness  of  the  plate], 
although  a  few  indicate  that  the  shearing  strength  of  the  rivet 
diminishes  after  the  intensity  of  the  bearing  pressure  exceeds 
a  certain  maximum  limit. 

34.  Theoretical  Deductions. 

Let  5  be  the  total  stress  at  a  riveted  joint ; 

/i>/2> /a>/4>  De  the  safe  tensile,  shearing,  compressive, 

and  bearing  unit  stresses,  respectively  ; 
/  be  the  thickness  of  a  plate,  and  w  its  width  , 


?2  THEORY  OF  STRUCTURES, 

N\*t  the  total  number  of  rivets  on  one  side  of  a  joint; 

it  be  the  total  number  of  rivets  in  one  row  ; 

p  be  the  pitch  of  the  rivets,  i.e.,  the  distance  centre  to 

centre  ; 

d  be  the  diameter  of  the  rivets  ; 
x  be  the  distance  between  the  centre  line  of  the  nearest 

row  of  rivets  and  the  edge  of  the  plate. 
Value  of  x.  —  It  has  been  found  that  the  minimum  safe 
value  of  x  is  d,  and  this  in  most  cases  gives  a  sufficient  overlap 
(—  2x\  while  x  =  \d  is  a  maximum  limit  which  amply  pro- 
vides for  the  bending  and  shearing  to  which  the  joint  may  be 
subjected.  Thus  the  overlap  will  vary  from  2d  to  ^d. 

x  may  be  supposed  to  consist  of  a  length  x^  to  resist  the 
shearing  action,  and  a  length  x^  to  resist  the  bending  action. 
It  is  impossible  to  determine  theoretically  the  exact  value  of 
JF,,  as  the  straining  at  the  joint  is  very  complex,  but  the  metal 
in  front  of  each  rivet  (the  rivets  at  the  ends  of  the  joint  ex- 
cepted)  may  be  likened  to  a  uniformly  loaded  beam  of  length 

d,  depth  x^  --  ,  and   breadth  /,  with  both  ends  fixed.     Its 

moment  of  resistance  is   therefore  -?t  Lr,  —  —j  ,  /  being  the 

maximum  unit  stress  due  to  the  bending.     Also,  if  P  is  the 
load  upon  the  rivet,  the  mean  of  the  bending  moments  at  the 

P 

end  and  centre  is  ^d. 

o 

Hence,  approximately, 


*  '-- 


It  will  be  assumed  that  the  shearing  strength  of  the  rivet 
is  equal  to  the  strength  of  a  beam  to  resist  cross-breaking. 
Single-riveted  lap  and  single-cover  joints  (Figs.  427  and  428). 

~~A  =  (p-d)tf^dtf^    .....     (I) 


THEORETICAL   DEDUCTIONS.  6/3 

*xjj[i  =  -—f%. 


d   ,    i      /      d'f,  .  . 

-••*•  =  2+ 4\/**-J7 '  '  (3) 

As  already  pointed  out,  these  joints  are  weakened  by  the 
bending  action  developed,  and  possibly  also  by  the  concentra- 
tion of  the  stress  towards  the  inner  faces  of  the  plates. 

Single-riveted  double-cover  joints  (Fig.  429). 


(4) 


tft  =  2—  /.. 

4 


(5) 


3rfr«    21  -2  4 


These  joints  are  much  stronger  than  joints  with  single 
covers.  Also,  equation  (4)  shows  that  the  bearing  unit  stress 
in  a  double-cover  joint  is  twice  as  great  (theoretically]  as  in  a 
single-cover  joint  (eq.  i),  so  that  rivets  of  a  larger  diameter 
may  be  employed  in  the  latter  than  is  possible  in  the  former, 

d 
for  corresponding  values  of  -. 


THEORY   OF  STRUCTURES. 

Chain-riveted  joints  (Fig.  431). 


5  =  N  --  /9  when  there  is  one  cover  only  ;     .     .     (8) 

5  =  N  -  /2  when  there  are  two  covers.    .     .     .     (9) 

This  class  of  joint  is  employed  for  the  flanges  of  bridge 
girders,  the  plates  being  piled  as  in  Figs.  436,  437,  438,  and  n 
being  usually  3,  4,  or  5. 

In  Fig.  437  the  plates  are  grouped  so  as  to  break  joint,  and 
opinions  differ  as  to  whether  this  arrangement  is  superior  to 
the  full  butt  shown  in  Fig.  438.  The  advantages  of  the  latter 


FIG.  436. 


FIG.  437.  FIG.  438. 

are  that  the  plates  may  be  cut  in  uniform  lengths,  and  the 
flanges  built  up  with  a  degree  of  accuracy  which  cannot  be 
otherwise  attained,  while  the  short  and  awkward  pieces  accom- 
panying broken  joints  are  dispensed  with. 

A  good  practical  rule,  and  one  saving  much  labor  and  ex- 
pense, is  to  make  the  lengths  of  the  plates,  bars,  etc.,  multiples 
of  the  pitch,  and  to  design  the  covers,  connections,  etc.,  so  as 
to  interfere  with  the  pitch  as  little  as  possible. 

The  distance  between  two  consecutive  joints  of  a  group 
(Fig.  437)  is  generally  made  equal  to  twice  the  pitch. 

An  excellent  plan  for  lap  and  single-cover  joints  is  to 
arrange  the  rivets  as  shown  in  Figs.  431  to  435. 

The  strength  of  the  plate  at  the  joint  is  only  weakened  by 
one  rivet-hole,  for  the  plate  cannot  tear  a^t  its  weakest  section, 


COVERS.  675 

i.e.,  along  the  central  row  of  rivets  (##),  until  the  rivets  be- 
tween it  and  the  edge  are  shorn  in  two. 

Let  there  be  m  rows  of  rivets,  i  1,22, 
3  3,  ...  (Fig.  439)- 

The    total    number   of    rivets    is   evi-  (_ 

dently  m\  O^O^O^O 

Let  /i  i  &  >  ?3 ,  £, , ...  be  the  unit  ten-  O,XO  I    ! 

sile  stresses  in  the  plate  along  the  lines  I  I,        

2  2>  3  3»  «  «  « »  respectively.     Then 


i 

+ 
U 


FIG.  439. 

5  =  (w  —  d}tf^  = nfft ,         for  the  line  i  i  ; 


n<?  t     »          \s 

(m  -  i)/;,  22; 

4 


3  3; 


44; 


Assume  that  /",  =  ^a .     Then  ze;  =  (n?  -\-  \]d. 

Hence,  by  substituting  this  value  of  w  in  the  first  of  the 

above  relations,  -  —  --  •£.    Since  qz ,  ^4 ,  .  .  .  are  each  less  than 
t        1 1  j  <i 

fi ,  the  assumption  is  justifiable. 

35.  Covers. — In  tension  joints  the  strength  of  the  covers 
must  not  be  less  than  that  of  the  plates  to  be  united.  Hence, 
a  single  cover  should  be  at  least  as  thick  as  a  single  plate  ;  and 
if  there  are  two  covers,  each  should  be  at  least  half  as  thick. 

When  two  covers  are  used  in  a  tension  pile  it  often 
happens  that  a  joint  occurs  in  the  top  or  bottom  plate,  so  that 
the  greater  portion  of  the  stress  in  that  plate  may  have  to  be 


676  THEORY  OF  STRUCTURES. 

borne  by  the  nearest  cover.     It  is,  therefore,  considered  advis- 
able to  make  its  thickness  five-eighths  that  of  the  plate. 

The  number  of  the  joints  should  be  reduced  to  a  minimum, 
as  the  introduction  of  covers  adds  a  large  percentage  to  the 
dead  weight  of  the  pile. 

Covers  might  be  wholly  dispensed  with  in  perfect-jump 
joints,  and  a  great  economy  of  material  effected,  if  the  dif- 
ficulty of  forming  such  joints  and  the  increased  cost  did  not 
render  them  impracticable.  Hence,  it  may  be  said  that  covers 
are  required  for  all  compression  joints,  and  that  they  must  be  as 
strong  as  the  plates  ;  for,  unless  the  plates  butt  closely,  the 
whole  of  the  thrust  will  be  transmitted  through  the  covers. 
In  some  of  the  best  examples  of  bridge  construction  the  ten- 
sion and  compression  joints  are  identical. 

36.  Efficiency  of  Riveted  Joints.*  —  The  efficiency  of  a 
riveted  joint  is  the  ratio  of  the  maximum  stress  which  can  be 
transmitted  to  the  plates  through  the  joint  to  the  strength  of 
the  solid  plates. 

Denote  this  maximum  efficiency  by  rf. 
Let  /  be  the  pitch  of  the  rivets  ; 

d       "       diameter  of  the  rivets  ; 

t       "       thickness  of  the  plates  ; 

ft      "       tenacity  of  the  solid  plate  ; 

mft  "  "         "     "  riveted  plate  ; 

fs      "       shearing  strength  of  the  rivets  ; 

N     "       number  of  rivets  in  a  pitch  length  ; 

e       "       ratio   of  the    strength   of   a   rivet   in  double 

shear  to  its  strength  in  single  shear. 
Then 

77,  =  efficiency  as  regards  the  plates  —  —  -  J  —  — 


eN-d'f, 
77,  =  efficiency  as  regards  the  rivets  —  -    -—7  —  .       (2) 


*  From  an  article  by  Professor  Nicolson  in  the  Engineer,  Oct.  9,  1888. 


EFFICIENCY  OF  RIVETED  JOINTS. 

The  efficiency  of  the  joint  is,  of  course,  the  smaller  of  these 
two  values  ;  and  the  joint  is  one  of  maximum  efficiency  when 
rji  =  %  =  rf  ;  that  is,  when 


m  -  --  = 


Ptft 
or 


In  this  expression  the  quantities  m  ft,  N,  and  e  are  con- 
stants for  any  given  joint,  being  of  necessity  known,  or  having 
been  fixed  beforehand  ;  and  the  equation  thus  expresses  one 
condition  governing  the  relations  of  the  three  variables  /,  d, 
and  /  to  each  other.  It  is  obvious,  however,  that,  in  order  to 
determine  the  values  of  any  two  of  these  variables  in  terms  of 
the  third,  another  relation  between  them  must  be  postulated. 
In  short,  in  designing  a  joint,  the  value  of  one  of  the  three 

ratios  --,  -,  and  —  must  be  fixed. 
at  t 

P 
CASE   I.    Suppose   that    the    ratio  -,  has  a   certain   value. 

This  is  very  frequently  the  quantity  predetermined  ;  but  it  is 
most  usually  done  by  fixing  the  value  of  77,  rj  very  obviously 

P  /         d\ 

involving  --   in  fact  y  —  m\i  —  -}. 

Equation  (3)  may  be  written 


4   t 
or 


If  the  ratio  —  b«  denoted  by  k,  then 


678  THEORY  OF  STRUCTURES. 

m(p  —  d) 

But  since  tj  —  — — -, 

P 


dm  —  rf'  *  ' 

Therefore,  substituting  in  (5), 

=  eN—k-^--:   ,  (7) 

m  —  rf  4    mft 

and,  ultimately, 

4     mft     r) 
^~eNn'J^'m~^ (8) 

The  process  of  designing  a  joint  of  maximum,  efficiency  for 
a  boiler  of  given   diameter  and   pressure   of  steam,   when    rj 


or  the  ratio        is  fixed,  is  then  as  follows:  Settle  the  number 
di 

of  rivets  per  pitch  (i.e.,  N)',  the  value  to  be  allowed  for  e  (de- 
pending on  the  nature  of  the  shearing  stress  on  the  rivets)  ; 
and  the  values  of  m,ft,  and  ff  Then  k  is  known  from  equa- 
tion (8). 

But  /  may  be  found  from  the  relation, 

pressure  X  diameter  =  rf  X  2tft  , 
or 

pressure  X  diameter 


Hence,  since  k  =  —  is   known,  d  may  be    found  ;   and   since 


d       m  — 


m  •      i      /*      i 

is  known,  p  is  also  fixed. 


CASE  II.    When    -,  the  ratio  of  rivet  pitch  to  plate  thick- 
ness, is  given,  equation  (5)  must  be  otherwise  manipulated. 


EFFICIENCY    OF  RIVETED  JOINTS.  679 

d 

Multiplying  it  by  —  ,  and  substituting  for  d  its  value  kt,  we 
have 


4    P     **ftP 

Putting  this  in  the  form  of  a  quadratic  equation  in  k, 

2 


For  brevity,  substituting  A  for  -^-  ,  T  for  ^-',  and  R  for  -, 
and  solving  the  quadratic, 


..     .     .    (12) 


The  method  of  designing  the  joint  is,  then,  as  follows  : 

A,  T,  and  R  being  known,  k  may  be  found  by  substituting 

their  values  in  equation  (12),  the  positive  sign  of  the  second 

term  being  taken. 
Now, 

(         d\  I         kt\  (         k\ 

r)  =  m(i  —  —  J  =  m\i  —  —  I  =  m\i  —  -=\  ; 

and  since  both  k  and  R  are  now  known,  the  thickness  of 
plate  (f)  may  be  found,  as  in  Case  I,  by  equation  (9).  The 
values  of  the  diameter  and  pitch  of  rivets  follow  at  once  from 
the  known  values  of  k  and  R. 

This  method  of  designing  a  joint  appears  to  be  the  most 
rational  of  the  three.  For  the  greatest  pitch  for  which  a  joint 
will  remain  steam-tight  depends  mainly  on  the  relation  of  pitch 
of  rivets  to  thickness  of  plates ;  although  it  is  also  affected  by 
the  relative  size  of  rivets  and  of  rivet-heads. 

CASE  III.  If  y,  or  k,  be  predetermined,  the  value  of  rj 
must  first  be  obtained,  in  order  that  the  plate  thickness  may 


680  THEORY  OF  STRUCTURES. 

be  found  by  means  of  equation  (9).     Now,  rj  =  m — - —  may 
be  put  into  the  form 

md 


m  — 


and  if  this  value  is  substituted  for/  in  equation  (4), 

md         leNn     f,          \ 

=  I k  — >  ~J-  I  la. 

m  —  rj       \  4       mft^t 

From  this  is  finally  deduced 


eNnkf 


The  plate  thickness  may  now  be  found  by  equation  (9)  ; 
the    diameter    of    rivet    from    d  =  kt,    and    the    pitch    from 

md 
p  =  -      —  .     In  the  above  investigations  no  account  has  been 

taken  of  the  effect  of  the  bearing  pressure  on  the  rivets   or 
plate. 

If  fc  be  the  allowable  bearing  pressure  per  projected  square 
inch  of  rivet  surface,  the  following  relation  must  obtain  : 


.    .    .    .    .    (14) 
This  may  be  written 


fc  ~         Nd 

Then  if  fe  be  estimated  by  this  equation,  and  if  it  should  1 
greater  than  43  tons  per  square  inch  in  a  lap  joint,  or  45  fr 
tons  in  a  butt  joint,  such  joint  will  fail  by  the  rivets  shec 
before  the  full  strength  of  the  plate  is  exerted,  as  Kent 
experiments  show  that  with  these  values  of  fc  the  rive 
not  attain  their  natural  ultimate  shearing  strength  (viz.,/5,, 
fail  at  shearing  stresses  much  below  this. 


EFFICIENCY   OF  RIVETED  JOINTS.  68  1 

Ar;ain,  the  maximum  allowable  ratio  of   —  (i.e.,  k]  as  the 

preliminary  datum  for  the  design  of  a  joint,  may  be  fixed  by 
using  the  expression 


<-> 


deduced  from  the  obvious  relation  —  similar  to  (14)  — 


eN-d*fs  =  Ndtfc. 
4 

win  suggests  the  relation  </=  £ 

~  Tin  designing  the  joint  by  any  of  the  methods  given  above, 
any  value  obtained  for  k  greater  than  that  supplied  by  (16) 
should  be  rejected. 


1  Note. — The  following  Tables  of  the  Weights  of  Bridges 
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bridges  in  question. 


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t^n-       r^ror-N      ooO^o 
M"^        ^^M         ^-^J^i 


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Q    2 

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688  THEORY   OF  STRUCTURES. 

TABLE   OF   LOADS    FOR    HIGHWAY    BRIDGES. 


Spaa  in  Feet. 

City  and  Suburban 
Bridges  liable  to 
Heavy  Traffic. 

Bridges  in  Manu- 
facturing Districts. 
Ballasted  Road. 

Bridges  in  Country 
Districts. 
Unballasted  Roads. 

joo  and  under 
100  to  200 
200  to  300 
300  to  400 
above  400 

100  Ibs.  per  sq.  ft. 

80    "      " 
70    "      " 
60    "      " 
50   "      " 

90  Ibs.  per  sq.  ft. 

60   "      " 
50   "      ••         " 
50   "      " 
50   "      " 

70  Ibs,  per  sq.  ft. 
60    "      " 
50   »      "        " 

45    "      "        " 

45    "      "        " 

f  I 


EXAMPLES.  089 


EXAMPLES. 


I.  A  bridge  of  N equal  spans  crosses  a  span  of  L  ft.;  the  weights  in 
tons  per  lineal  foot  of  the  main  girders  of  the  platform,  permanent  way, 
etc.,  and  of  the  live  load,  are  iu\ ,  w* ,  ws ,  respectively.  Show  that 

LA 


where          A  =  w^pk  +  r)  +  iv*(pk  +  q)    and    B  =  pk  +  r, 

k  being  the  ratio  of  span  to  depth,  and  p,  g,  r  numerical  coefficients. 
Hence  also  determine  the  limiting  span  of  a  girder. 

If  X  is  the  cost  of  a  pier  and  if  Y  is  the  cost  per  ton  of  the  super- 
structure, find  the  value  of  7V  which  will  make  the  total  cost  per  lineal 
foot  a  maximum,  and  prove  that  this  is  approximately  the  case  when  the 
spans  are  so  arranged  that  the  cost  of  one  span  of  the  bridge  structure 
is  equal  to  the  cost  of  a  pier. 

Ans.  A  span  <  — - —     -    Cost  is  a  minimum  when  N—  LB  =  Z4/fU_, 

and  the  minimum  cost  of  the  span  =  X(  i  —  —  j  =  X,  approx. 

2.  A  car  of  weight  W  for  a  gauge  of  4  ft.  8£  in.  is  33  ft.  long,  6  ft.  deep, 
and  its  bottom  is  2  ft.  6  in.  above  the  rails.     Find  the  additional  weight 
thrown  upon  the  leeward  rail  when  the  wind  blows  upon  a  side  of  the 
car  with  a  pressure  of  20  Ibs.  per  square  foot.     Also  find  the   minimum 
pressure  that  will  blow  the  car  over.  Ans.  4625.84  Ibs. ;  .428  W. 

3.  A  lattice  girder  200  ft.  long  and  20  ft.  deep,  with  two   systems  of 
right-angled    triangles,  carries  a  dead  load  of  800   Ibs.  per  lineal    foot. 
Determine  the  greatest  stresses  in  the  diagonals  and  chords  of  the  fourth 
bay  from  one  end  when  a  live  load  of  1200  Ibs.  per  lineal  foot  passes  over 
the  girder. 

Ans.  If  riveted :  Diagonal  stress  =  37,2004/2  Ibs. ; 

Chord  stress  =  450,000  Ibs. 

If  pin-connected :  Diagonal  stress  =  44,800  4/2  and  29,600  4/2  Ibs.; 
Chord  stress  =  460.000  Ibs. 


690  THEORY  OF  STRUCTURES. 

4.  A  lattice-girder  80  ft.  long  and  8  ft.  deep  carries  a  uniformly  dis- 
tributed load  of  144,000  Ibs.     Find  the  flange  inch-stresses  at  the  centre, 
the  sectional  area  of  the  top  flange  being  56^  sq.  in.  gross,  and  of  the 
bottom  flange  45  sq.  in.  net. 

What  should  be  the  camber  of  the  girder,  and  what  extra  length 
should  be  given  to  the  top  flange,  so  that  the  bottom  flange  of  the  loaded 
girder  may  be  truly  horizontal  ?     (E  =  29,000,000  Ibs.) 
Ans.  3185.8  Ibs.  ;  4000  Ibs. 

•ri  =  -29735  :  **  =  -2987  J  S*  ~  s*  =  Ttffifr. 

5.  A  lattice-girder  80  ft.  long  and   10  ft.  deep,  with  four   systems  of 
right-angled  triangles,  carries  a  dead  load  of  1000  Ibs.  per  lineal  foot. 
Determine  the  greatest  stresses  in  the  diagonals  met  by  a  vertical  plane 
in  the  seventh  bay  from  one  end  when  a  live  load  of  2500  Ibs.  per  lineal 
foot  passes  over  the  girder.     Design  the  flanges,  which  are  to  consist  of 
plates  riveted  together. 

The  lattice-bars  are  riveted  to  angle-irons.  Find  the  number  of  f-iri, 
rivets,  required  to  connect  the  angle-irons  with  the  flanges  in  the  first 
bay,  10,000  Ibs.  per  square  inch  being  the  safe  shearing  strength  of  the 
rivets. 

Ans.  \iriveted:  Diagonal  stress  =  10,664^1/2  Ibs. 

If  pin-connected  ':       "  =  9062^4/2;  62501/2;  15,4681/2; 

11,8751/2  Ibs. 
22  rivets 


6.  The  bracing  of  a  lattice-girder  consists  of  a  single  system  of  tri- 
angles in  which  one  of  the  sides  is  a  strut  and  the  other  a  tie  inclined  to 
the  horizontal  at  angles  of  a  and  ft  respectively  ;  in  order  to  give  the 
strut  sufficient   rigidity  its  section  is  made  k  times  that  indicated  by 
theory,  the  coefficient  k  being  >  unity.     Show  that  the  amount  of  ma- 
terial in  the  struts  and  ties  is  a  minimum  when 

tan  a  =  k  tan  ft. 

7.  A  lattice-girder  of  40  ft.  span,  5  ft.  depth,  and   with   horizontal 
chords  has  a  web  composed  of  two  systems  of  right-angled  triangles  and 
is  designed  to  support  a  dead  and  a  live  load,  each  of  -J-  ton  per   lineal 
foot.     Determine  the    maximum    stresses  in  the  members  of  the  third 
bay  from  one  end  met  by  a  vertical  plane. 

Ans.  \iriveted:  Diagonal  stress  =  V//2  tons  J 

Chord  stress  =  27  tons. 

If  pin-connected  :  Diagonal  stress  =  ff  1/2  and  f£  ^2  tons  ; 
Chord  stress  =  26  tons. 

8.  A  lattice-truss  of  100  ft.  span  and  10  ft.  depth  has  a  web  composed 
of  four  systems  of  right-angled  triangles.     The  maximum  stress  in  the 


EXAMPLES.  691 

diagonal  joining  the  sixth  apex  in  the  upper  chord  to  the  fourth  apex  in 
the  lower  is  16  tons.  Find  the  dead  load,  the  live  load  being  i  ton  per 
lineal  foot,  assuming  the  truss  to  be  (a)  riveted,  (b)  pin-connected. 

Ans.   (a)  .554  ton  ;  (b)  1.062  tons. 

9.  A  lattice-girder  of  40  ft.  span  has  a  web  composed  of  two  systems 
of  triangles  (base  =  10  ft.)  and  is  designed  to  carry  a  live  load  of  1600 
Ibs.  per  lineal  foot  and  a  dead  load  of  1200  Ibs.  per  lineal  foot.     Defin- 
ing the  stress-length  of  a  member  to  be  the  product  of  its  length  into 
the  stress  to  which  it  is  subjected   find  the  depth  of  the  truss  so  that  its 
total  stress-length  may  be  a  minimum.  Ans.   10.19  ft- 

10.  Determine  the  maximum  stresses  in  the  members  of  a  lattice- 
truss  of  40  ft.  span  and  4  ft.  depth,  with  two  systems  of  triangles  (base 
=  8  ft.),  (a)  when  riveted  together;  (b)  when  pin-connected.    Dead  load 
=  J  ton  per  lineal  foot,  live  load  =  $  ton  per  lineal  foot. 

Ans.  Bays—  ist;  2d  ;                3d;              4th;  5th. 

(a)  Diags.        6f  4/2~;  5.534/2";  4.054/2";  2.854/2";  if  4/2~tons  ; 

Tens.chd.      6|  ;  i8f;               2;f;               33!;  36$. 

Comp.  chd.  Same. 


(  7.5^2  p.rVS       4.7^2  j     2.34/1  v\  2.34/2^ 

Tens.chd.      6;  18;  27;  33;  36  " 

Comp.  chd.  Same. 

ii.  The  platform  of  a  single-track  bridge  is  supported  upon  the  top 
chords  of  two  Warren  girders  ;  each  girder  is  100  ft.  long,  and  its  brac- 
ing is  formed  of  ten  equilateral  tfiangles  (base  10  ft.);  the  dead  weight 
of  the  bridge  is  900  Ibs.  per  lineal  foot.  ;  the  greatest  total  stress  in  the 
seventh  sloping  member  from  one  end  when  a  train  crosses  the  bridge 
is  41,394.8  Ibs.  Determine  the  weight  of  the  live  load  per  lineal  foot. 
Prepare  a  table  showing  the  greatest  stress  in  each  bar  and  bay  when  a 
single  load  of  1  5,000  Ibs.  crosses  the  girder. 

Ans.         t\    tz   >t3    h    ft 

WWWXAAA/         277H  Ibs.  per  lin.  ft. 

FIG.  440. 

Stresses  in  diagonals  :       </,  =  of,  =  9  4/J;    da  —  d*  =  8  1/3  ; 

d*  =  d«  =  7  |/5  ;    <*,  =  //„  =  6  j/3  ; 

d*  =d™=  5  1/3  tons. 
Stresses  in  compression  :  c\  =  9  1/3  ;  c<i  =  16  4/3";  c»  =21  4/3"; 

<r4  =  24  1/3  tons. 
Stresses  in  torsion  :  /i  =  4i  I/J  :  /a  =  1  2  |/J;  /3  =  1  7^  1/3  ; 

/4  =  21  4/3  ;  /5  =  22£  4/3  tons. 


692 


THEORY  OF  STRUCTURES. 


12.  A  Warren  girder  with  its  bracing  formed  of  nine  equilateral  tri- 
angles (base  =  10  ft.)  is  90  ft.  long,  and  its  dead  weight  is  500  Ibs.  per 
lineal  foot.  Determine  the  maximum  stresses  in  each  member  when 
a  live  load  of  1350  Ibs.  per  lineal  foot,  preceded  by  a  concentrated 
load  of  18,000  Ibs.,  crosses  the  girder,  assuming  that  every  joint  is  loaded. 
The  diagonals  and  verticals  are  riveted  to  angle-irons  forming  part  of 
the  flanges. 

How  many  f-in.  rivets  are  required  for  the  connection  of  the  several 
members  meeting  at  the  third  apex  in  the  upper  chord?  (23,  6,  and  13.) 
How  many  are  required  in  the  first  bay  of  each  chord  to  prevent  longi- 
tudinal slip?  (15  in  tension  chord  and  18  in  compression  chord.) 


Ans. 


FIG.  441 

Tension  chord  stresses  : 


t9  =/,„  = 


V3 
333125 

V3 
415125 


V3 
39462 5 


164000          287000          -360000 

Compression  chord  stresses:   c\  =  - — ;=--;  c*— — — -  ;  ca=      — 

V3  V3  V3 

410000 

V~3 

178500  159500          141250 

Stresses  in  sloping  members :  di=  — ;=-  ;  d*= — 7=-  ;  as= — ^=—  ; 

T   3  r    3  T  3 

I2375o  107000  91000 

^3~  ^3  "4/3" 


70250 


47500 


Ibs. 


__^d   _  I075Q 

^3"  VJ   '  4/3" 

The  stresses  ^/jo,  d\\ ,  da,  are 


max.  stresses  of  an   opposite  kind 
to  those  due  to  dead  load. 

Verticals  :    Max.  load  on  each  vertical  =  20,  500  Ibs. 
13.  If  a  force  of  5000  Ibs.  strike  the  bottom  chord  of  the  girder  in  the 
preceding  question  at  20  ft.  from  one  end  and  in  a  direction  inclined  at 
30°  to  the  horizontal,  determine  its  effect  upon  the  several  members. 


EXAMPLES.  693 

14.  A  Warren  girder  for  a  single-track  railway  bridge  consists  of 
eight  equilateral  triangles  and  has  to  cross  a  span  of  96  ft.  ;  the  platform 
is  on  the  bottom  chord  ;  the  loads  per  lineal  foot  for  which  the  truss  is  to 
be  designed  are  2250  Ibs.  due  to  engine,  1  500  Ibs.  due  to  train,  and  450  Ibs. 
due  to  bridge.     Determine  the  maximum  stresses  (both  tensile  and  com- 
pressi  ve)  in  the  members  met  by  vertical  planes  immediately  on  the  right  of 
the  second,  third,  and  fourth  apices  in  the  compression  chord.     Also, 
find  how  many  |-in.  rivets  are  required  to  connect  the  diagonals  met  by 
these  planes  with  the  chords  and  to  prevent  any  tendency  to  longitudinal 
slip  between  the  support  and  the  first  apex,  and  between  the  first  and 
second  apices  in  the  tension  chord. 

15.  The  accompanying  figure  represents  the  half-truss  fora  bridge  of 
80  ft.  span.     Show  how  to  determine  the  stresses 

in  the  several  members.     Depth  at  centre  =  12  ft.; 
at  £=12  ft.;  at  .4  =  6  ft. 


FIG.  442. 

16.  A  Warren  girder  composed  of  eight  equilateral  triangles  has  its 
upper  chord  in  tension  and  has  every  joint  loaded  with  a  weight  of  2  tons, 
the  loads  being  transmitted  to  the  joints  in  the  lower  chord  by  means  of 
vertical  struts.     The  span  =  80  ft.     Find  the  stresses  in  all  the  members. 

Ans.  Bays  in  tension  chord  :  ist  =   5  4/3  ;  2d  ==  13  4/3  ; 

3d  =   i8i  4/3  ;  4th  =  21  4/3  tons. 
Bays  in  compression  chord  :  ist  =  9i  1/3  ;    2d  =  16  4/3  ; 

3d  =  20  4/3  ;  4th  =  21^  4/3  tons. 
Stresses  in  verticals  :    In  each  vertical  =  2  tons. 
Stresses  in  diagonals  :  ist  =  10  4/3  ;  2d  =  8f  4/3 

3d  -.=  7*4/3;  4th  =64/3; 
5th  =  4f  4/3  ;  6th  =  3fc  4/3  ; 
7th  =  24/3;    8th  =£  4/3  tons. 

17.  A  Warren  girder,  with  the  platform  on  the  lower  boom,  carries  a 
load  of  20  tons  at  the  centre.     Find  the  stress  in  each  member,  and  also 
find  the  weight  at  each  joint  of  lower  boom  which  will  give  the  same 
stresses  in  the  centre  bays. 

There  are  six  bays  in  the  lower  chord. 

Ans.   Stress  in  each  diagonal  =  2¥°  4/3  tons. 

Tens,  chord  :  stress  in  ist  bay  =  y>  4/3  ;  2d  =  10  4/3  ;    3d  —  5/  -^3 

tons. 
Cornp.  chord  :  stress  in  ist  bay  =2F°  4/3  ;  2d  =  4?°  4/3  ;    3d  =  20  ^3 

tons. 
Weight  at  each  joint  =  5{f  tons. 


694  THEORY  OF  STRUCTURES. 

1 8.  The  accompanying  truss  of  240  ft.  span  and  30  ft.  deep  is  to  be  de- 
signed for  a  panel  engine  load  of  24,000  Ibs.,  a  panel  train  load  of  18,000 

i  Ibs.,  and   a   panel   bridge   load   of   12,000 

Ibs,     Determine  graphically  the  maximum 
//   stresses  in  the  members  met  by  the  ver- 
tical  MN.     Also,  draw  a  stress  diagram 
for  the  whole   truss  when   it   is   covered 
FIG.  443.  with  a  uniformly  distributed  live  load  of 

180,000  Ibs. 

19.  Loads  of  3f ,  6,  6,  6,  and  6  tons  follow  each  other  in  order  over  a  ten- 
panel  truss  at  distances  of  8,  5f,  4^,  and  4!  ft.  apart.    Determine  the  posi- 
tion of  the  loads  which  will  give  the  maximum   diagonal  and  chord 
stresses  in  the  third  and  fourth  panels.     Span  =  120  ft. 

20.  Determine  the  moment  of  resistance  of  a  floor-beam  for  the  Sault 
Ste.  Marie  Bridge  from  the  following  data :  Floor-beams,  16'  6"   long 
and   23'  lof"  apart ;  the  dead  weight  of    the  flooring,  stringers,  etc. 
=  800  Ibs.  per  lineal  foot  of  floor-beam;  the  live  load  as  given  in  Fig.- 
407,  Art.  20,  page  639 ;  the  load  is  transmitted  to  the  floor-beam  by  four 
lines  of  stringers  so  spaced  as  to  throw  two-thirds  of  the  load  upon  the 
inner  pair,  which  are  3  ft.  C.  to  C. 

21.  In  a  truss-bridge  the  panels  are  17  ft.  and  the  floor-beams  13  ft. 
in  length.     Loads  of  8,  12,  12,  12,  12,  10,  10,  10,  and  10  tons  follow  each 
other  in  order  over  the  bridge  at  the  distances  of  7^,  4^,  4^,  4!,  7^,  5^,  6£, 
and  5^   ft.  apart.     Determine  the  moment  of  resistance  of  the  beam, 
taking  the  load  due  to  the  platform,  etc.,  to  be  500  Ibs.  per  lineal  foot. 

22.  If  the  bridge  in  the  preceding  question  is  of  the  riveted  type  with 
a  single  diagonal  system,  and  with  verticals  at  the  panel  points,  the  num- 
ber of  the  panels  being  ten,  find  how  many  i-in.  rivets  are  required  in 
the  third  panel  from  one  end  to  connect  the  web  with  the  chords,  assum- 
ing the  panel  live  load  to  be  30,000  Ibs.  and  the  panel  dead  load  to  be 
10,000  Ibs. 

23.  With  the  loading  given  by  Fig.  409,  Art.  20,  page  639,  design  a 
floor-beam  for  a  single-track  bridge  with  panels  22  ft.  long,  the  weight  of 
the  platform  being  450  Ibs.  per  square  yard,  and  of  each  longitudinal  200 
Ibs.  per  lineal  yard. 

24.  Prepare  a  table  giving  the  stresses  of  the  several  members  of  a 
double-intersection  through-truss  of  154  ft.  span,  20  ft.  depth,  and  with 
eleven  panels.     The   panel   engine,  live,  and   bridge   loads   are  56,000, 
35,000,  and  16,800  Ibs.,  respectively. 

25.  Prepare  a  table  giving  the  stresses  in  the  several  members  of  a 
double-intersection  deck-truss  of  342  ft.  span,  33  ft.  depth,  and  with 
eighteen    panels.     (Double-track    bridge.)      The    panel    engine,    train 
(or  live),  and  dead  loads  are  96,000,  54,000,  and  36,200  Ibs.,  respectively. 

26.  Prepare  a  table  giving  the  stresses  in  the  several  members  of  a 


.EXAMPLES. 


.695 


through-truss  for  a  double-track  bridge  of  342  ft.  span,  40  ft.  depth,  and 
with  nineteen  panels.  The  panel  engine,  live,  and  dead  loads  are  96,000, 
53,000,  and  43,200  Ibs.,  respectively  (double-intersection). 


Cl  C2   C3  C4   C5  C6   C7  C8  C9 


Ans. 


FIG.  444. 


Chord 
Panel. 

Mult. 

7326 

Mult. 

5063 

Total  Shear 
transmitted 
to  Panel. 

Tan- 
gent. 

Panel  Stress 
from  Shear. 

Total 
Panel 

Stress. 

t   =  t. 

18 

131868 

i53 

774639 

906507 

•45 

407929 

407929 

t^ 

— 

-  7326 

8oi 

407571* 

400245^ 

•45 

180110.475 

588039 

t^ 

— 

72* 

•9 

*6 

— 

6ij- 

•9 

— 

53* 

•  9 

2 



42* 

•9 

t6 

— 

-   7326 

34* 

1  74673* 

167347* 

•9 

150612$ 

tg                    — 

23* 

•  9 

'10        j     -. 

•9 

( 

jeo 

•45 

1 

c\ 

i 

1 

•< 

80* 

•45 

V 

921696 

1 

72* 

•9 

| 

Cn 

_ 

61* 

•9 

ca 

— 

53* 

•9 

C4 

— 

42* 

•9 

f| 

— 

34* 

•  9 

C6 

— 

23* 

•9 

C-r 

_ 

15* 

•9 

c 



4* 

•9 

'* 

— 

-   3* 

•9 

1 

Shear 

Diag. 

Mult. 

5°53 

Mult. 

2790 

due  to 
Live 

Mult. 

2274 

Total 
Shear. 

Secant 

Max. 
Stress. 

Load. 

p 

18 

153 

171 

.0965 

4t 

J7 

63* 

.0965 

^2 

16 

56* 

72* 

•345 

\ 

i5 

75795 
70742 

48* 
42* 

118575 

2III  IO 
189317 

53* 

139851 
121659 

350961 
310976 

•345 
•345 

12 

65689 
60636 

35* 
30* 

99045 
85095 

164734 

42* 
34* 

96645 
78453 

261379 
224184 

•345 
•345 

301528 

di 

II 

24* 

23* 

•345 

dg 

10 

20^ 

15* 

•345 

dg 

9 

4* 

•345 

^10 

8 

40424 

I2* 

34875 

75299 

-  3* 

-  7959 

•345 

90572 

a'll 

7 

35371 

8* 

-14* 

-  32973 

•345 

35122 

/1  =  139200  Ibs. 

>2   =   350961     " 
/3   =   310976     " 


£'4  =  261379  Ibs. 

Z'5  =  224184  " 
"6  =  177377  " 


=  142972  Ibs. 

=   98955  " 
=   67340  " 


27.  Prepare  a  table  showing  the   stresses  in  the   several    members' 
(including  counters)  of  a  ten-panel  double-track  through  railway  bridge 
of  184^  ft.  span  and  34  ft.  depth,  the  live  and  dead  loads  being  respect- 
ively 2250  Ibs.  and  1100  Ibs.  per  lineal  foot.   (Thamesville  Bridge.) 

28.  Determine  the  minimum  stress-length  (stress  in  a  member  multi- 
plied by  its  length)  for  a  double-intersection  Pratt  truss  of  154  ft.  span  and 
with  eleven  panels.     The  panel  loads  for  engine  =  44,000  Ibs.,  for  train  = 


696  THEORY  OF  STRUCTURES. 

27,500  Ibs.,  for  bridge  =  13,200  Ibs.  ;  coefficient  of  working  strength  =  8000 
Ibs.  per  square  inch  for  both  compression  and  tension. 

29.  A  six-panel  single-intersection  Pratt  truss  is  uniformly  loaded. 
Assuming  the  same  coefficient  of  strength  both  for  compression  and 
tension,  show  that  the  economy  of  material  will  be  greatest  when  the 
diagonals  are  inclined  at  32°  25' to  the  vertical. 

30.  A  double-intersection  truss  for  a  single-track  through-bridge  of 
204  ft.  span  is  29  ft.  deep,  20  ft.  wide,  and  has  twelve  panels.     Find  the 
stresses  produced  in  the  members  of  the  leeward  truss  by  a  panel  wind- 
pressure  of  5000  Ibs.  acting  8  ft.  above  base  of  rails  (5-ft.  gauge). 

Ans.  Sloping  members  :       ist  =  27500  sec  oc ;       2d  =  12708$  sec  a  ; 

3d  =  10208$  sec  ft ;  4th  =  7708$  sec  ft\ 
5th  =  5208$  sec  a  ;  6th  =  2708$  sec  0 ; 
7th  =  208$  sec  ft. 

Tension  chord  :  ist  panel  =  27500  tan  a  =  2d  ;  3d  =  40208$ tan  a; 
4th  =  40208$  tan  a  +  10208$  tan  ft; 
5th  =  40208$  tan  a  +  17916!  tan  ft ; 
6th  =  40208$  tan  a  +  23125  tan  ft. 

Compression  chord  :    ist  =  40208$  tan  a  +  10208$  tan  ft ; 
2d  =  40208$  tan  a  +  17916$  tan  ft  ; 
3d  =  40208$  tan  n  +  23125  tan  ft  ; 
4th  =  40208$  tan  a  +  25833$  tan  ft ; 
5th  =  40208$  tan  a  +  26041!  tan  ft. 
Verticals :  ist  =  5000 ;      2d  =  7708$ ;   3d  =  5208$; 

4th  =  2708$  ;  5th  =  208$  Ibs. 
tan  a  =  $| ;      tan  ft  =  ff . 

31.  In  the  preceding  question  find  the    maximum    stresses  in   the 
members  of  the  fourth  panel  met  by  a  vertical  plane ;  engine  panel  load 
=  85,000  Ibs.,  train  panel  load  =  40,800  Ibs.,  bridge  panel  load  —  22,500 
Ibs. 

Ans.  Stresses  in  tension  chord  =  456^430.45  Ibs. ;  in  compression 
chord  —  645,311.77  Ibs.  ;  in  sloping  members  =  206,242.5  Ibs.; 
and  139,705.62  Ibs. 

32.  Each  of  the  two  Pratt  single-intersection  five-panel  trusses  for  a 
single-track   bridge  is  55  ft.  centre  to  centre  of  end  pins  and  1 1  ft.  6  in. 
deep.     Timber  floor-beams  are  laid  upon  the  upper  chords  2f  ft.  centre 
to  centre  ;  the  width  between  the  chords  =  10  ft.     Find  the  proper  scant- 
ling of  the  floor-beams   for  the  loading  given    in    Fig.  407,  page  639. 
Also  determine  the  maximum  chord  and  diagonal  stresses  in  the  centre 
panel  due  to  the  same  live  load. 

33.  Prepare  a  table  giving  the  stresses  in  the  several  members  of  a 
double-intersection  deck-truss  of    342  ft.  span,  40  ft.  depth,  and  with 


EXAMPLES.  697 

nineteen  panels.     (Double-track  bridge.)     The  panel   engine,  train  (or 
live),  and  dead  loads  are  96,000,  53,000,  and  43,200  Ibs.,  respectively. 

34.  Prepare  a  table  giving  the  stresses  in  the  several  members  of  a 
deck-truss  for  a  double-track   bridge  of  342  ft.  span,  33   ft.  depth,  and 
with  eighteen  panels.    The  panel  engine,  live,  arid  dead  loads  are  96,000, 
54,000,  and  36,000  Ibs.,  respectively. 

35.  The  two  trusses  for  a  16  ft.  roadway  are  each  100  ft.  in  the  clear, 
17  ft.  3  in.  deep,  and  of  the,  type   repre- 
sented in  the  figure  ;  under  a  live  load  of 

1 1 20  Ibs.  per  lineal  foot  the  greatest  total 
stress  in  AB  is  35,400  Ibs.  Determine  the 
permanent  load.  FIG.  445. 

The  diagonals  and  verticals  are  riveted  to  angle-irons  forming 
part  of  the  flanges.  How  many  f-in.  rivets  are  required  for  the  con- 
nection of  AB  and  BC  at  B  ?  Also,  how  many  are  required  between 
A  and  C  to  resist  the  tendency  of  the  angle-irons  to  slip  longitudinally  ? 
Working-shear  stress  =  10,000  Ibs.  per  square  inch. 

Ans.  708.9  Ibs. ;  8;  4;   n. 

36.  The   compression  chord  of  a  bowstring  truss  is  a  circular  arc  of 
80  ft.  span  and  10  ft.  rise ;  the  bracing  is  of  the  isosceles  type,  the  bases 
of  the  isosceles  triangles  dividing  the  tension  chord  into  eight  equal 
lengths.     Determine  the  maximum  stresses  in  the  members  met  by  a 
vertical  plane  28  ft.  from  one  end.     The  live  and  dead  loads  are   each 
i  ton  per  lineal  foot. 

37.  Design  a  parabolic  bowstring  truss  of  80  ft.  span  and  10  ft.  rise 
for  a  dead  load  of  £  ton  and  a  live  load  of   i  ton  per  lineal  foot.     The 
joints  between  the  web  and  the  tension  chord  are  to  divide   the  latter 
into  eight  equal  divisions. 

38.  The  compression  chord  of  a  bowstring  truss   is  a  circular  arc. 
The  depth  of  the  truss  is  14  ft.  at  the  centre  and  5  ft.  at  each  end  ;  the 
span  =  100  ft.  ;  the  load  upon  the  truss  =  840  Ibs.  per  lineal  foot.     Find 
the  stresses  in  all  the  members.     Determine  also  the  maximum  stresses 
in  the  members  met  by  a  vertical  25  ft.  from  one  end  when  a  live  load 
of  looo  Ibs.  per  lineal  foot  crosses  the  girder.     What  counter-braces  are 
required  ? 

39.  A  Pratt  truss  with  sloping  end  posts  has  a  length  of  150  ft.  centre 
to  centre,  and  a  height  of  30  ft.  centre  to  centre,  with  panels  15  ft.  long ; 
the  dead    load  is  3000  Ibs.  per  lineal  foot,  and  the  live  load    1200  Ibs. 
Determine  the  maximum  stresses  in  the  end  posts,  in  the  third  post  from 
one  end,  in  the  middle  of  the  bottom  chord,  and  in  the  members  of  the 
third  panel  met  by  a  vertical  plane. 

40.  Design  a  cross-tie  for  a  double-track  open-web  bridge,  the  ties 


698  THEORY  OF  STRUCTURES. 

being  18  ft.  5  in.  centre  to  centre,  and  the  live  load  for  the  floor  system 
being  8000  Ibs.  per  lineal  foot. 

41.  A  bowstring  roof-truss  of  50  ft.  span,  15  ft.  rise,  and  five  panels  is 
to  be  designed  to  resist  a  wind   blowing  horizontally  with  a  pressure  of 
40  Ibs.  per  square  foot*    The  depth  of  the  truss  at  the  centre  is  10  ft. 
Determine,  graphically,  the  stresses  in  the  several  members  of  the  truss, 
assuming  that  the  roof  rests  on  rollers  at  the  windward  support. 

42.  A  bowstring  truss  of  120  ft.  span  and  15  ft.  rise  is  of  the  isosceles 
braced  type,  the  bases  of   the  isosceles  triangles  dividing  the  tension 
chord  into  twelve  equal  divisions  ;  the  dead  and  live  loads  are  £  ton  and 
i  ton  per  lineal  foot,  respectively.     Find  the  maximum  stresses   in  the 
members  met  by  vertical  planes  immediately  on  the  right  of  the  second 
and  fourth  joints  in  the  tension  chord. 

43.  The  figure  is  a  skeleton    diagram  of  the  Sault  Ste.  Marie  Bridge 
(C  P.  R.).     Span  =  239  ft. ;  there  are  ten  panels,  each  of  23.9  ft.,,  say  24 

493000499,000 


275,000   275,000   416,000    478,000   484,000 

FIG.  446. 

ft. ;  the  length  of  the  end  verticals  =  27  ft.,  of  the  centre  verticals  =  40 
ft.;  width  on  truss  centres  =  17^  ft.  The  bridge  is  designed  to  bear  the 
loading  given  by  Fig.  407,  page  639.  Show  that — 

(a)  The  stresses  in  every  panel  length  of  each  chord  are  greatest  when 
third  driver  is  at  a  panel  point ;  and  find  the  value  of  the    several 
stresses. 

:  (b}  The  stresses  in   the  verticals  a  and  the  diagonals  b  are  greatest 
when  the  third  driver  is  at  a  panel  point ;  and  find  their  values. 

/  (f)  The  stresses  in  the  remaining  members  of  the  truss   are  greatest 
when  the  second  driver  is  at  a  panel  point  ;  and  find  their  values. 

(d)  The  maximum  stresses  in  the  verticals  a!  vary  from  a  tension  of 
64,000  Ibs.  to  a  compression  of  11,000  Ibs. 

(e)  The  stress  in  the  counter-brace  c  is  nil. 

Am.  The  values  of  thestresses  in  the  several  members  are  marked 
on  the  diagram.  They  are  deduced  from  the  distributions 
given  in  the  table  on  page  642,  and  are  correct  within  a  very 
small  percentage. 

44.  The  figure  represents  a  counterbalanced  swing-bridge,  16  ft.  deep 
and  wholly  supported  upon   the  turn-table  at  A 
and  B ';  the  dead  weight  is  650  Ibs.  per  lineal  foot 
of  bridge ;  the  counterpoise  is  hung  from  C  and  D. 
FIG.  447?  Find  its  weight,  assuming  (a)  that  the  whole  of  it 

is  transmitted  to  B ;  (b)  that  a  portion  of  it  is  transmitted  to  A  through 


EXAMPLES. 


699 


a  member  BE,  sufficient  to  make  the  reactions  at  A  and  B  equal. 
determine  the  stresses  in  the  several  members  of  the  truss. 


Also, 


Ans.  Counterpoise  in  case  (a)  =  26,162$  Ibs. ; 
in  case  (b)  =  22,186$%  Ibs. 
Stress  transmitted  through  BE  in  case  (£)=  24,012  Ibs. 

45.  The  figure  represents  a  counterbalanced  swing-bridge ;  the  dead 
load  upon  the  bridge  is  650  Ibs.  per  lineal 
foot ;  the  counterpoise  is  suspended  from 
CD.  Find  its  value,  the  joint  at  E  being 
so  designed  that  the  whole  of  the  load 
upon  the  bridge  is  always  transmitted  FIG.  44b. 

through  the  main  posts  EA,  EB,  and  is  evenly  distributed  between  the 
points  of  support  at  A  and  B. 

Ans.  20,694.3  Ibs. 

46.  Find  the  stresses  in  the  several  members  of  the  truss  in  the  pre- 
ceding question  (a)  when  the   bridge  is  open  ;  (b)  when  the  bridge  is 
closed    and   is   subjected  to  a  live    load  of   3000  Ibs.   per  lineal  foot. 
Height  of  truss  at  E  =  16  ft.,  at  F  =  8  ft. 

47.  Prepare  a  table  giving  the  stresses  in  the  several  members  of  a 
a   cs  c3  C4  C6 .         single-intersection  through-truss  of 


y\  \2N3J\4JX5  X6  \  \  \  \  J  54  ft- span>  20  ft-  dePth' and  with 

/  fi\ff2\f3\ffAfo\j  \J    \j  \J  \J  \    eleven  panels.     The  panel  engine, 

IQ     U&    £3       14       tfl      1C 

FIG.  449.  live,  and  dead  (or  bridge)  loads 

27,500,  17,600,  and  8470  Ibs.,  respectively. 

Ans. 


Diag. 

Mult. 

2500 

Mult. 

1600 

Sum. 

Mult. 

770 

Sum. 

sec. 

Total  M% 
Stress.  Y 

p 

IO 

25000 

45 

72000 

97000 

55 

42350 

139350 

22 

170007 

dl 

9 

22500 

36 

57600 

80100 

44 

33«80 

113980 

22 

139056 

dn 

g 

200  oo 

28 

44800 

64800 

33 

25410 

90210 

22 

110057 

<^3 

7 

17500 

21 

33600 

51100 

22 

16940 

68040 

22 

83009 

6 

15000 

15 

24000 

39000 

11 

8470 

47470 

22 

579J4 

</6 

5 

12500 

IO 

16000 

28500 

0 

28500 

22 

34770 

*6 

4 

10000 

6 

9600 

19600 

—  II 

-8470 

"130 

22 

13579 

Panel. 

Mult. 

3270 

Mult. 

2370 

Sum. 

tan. 

Panel 
Stress. 

Total  Panel 
Stress. 

'i  =  t* 
'3 
'4 
'5 

*6 

10 

—  I 
—  I 

—  I 
—  I 

32700 
-  3270 
-  3270 
—  3270 
-  3270 

45 
45 
34 
23 

12 

106650 
106650 
80580 

54510 
28440 

139350 
103380 
77310 
51240 
25170 

f? 

97545 
72366 

54"7 
35868 
17619 

97545 
169911 
224028 
259896 
2775*5 

700 


THEORY  OF  STRUCTURES. 


Panel. 

Mult. 

3270 

Muit. 

2370 

Sum. 

tan. 

Total 
Stress. 

Total  Max. 
Stress. 

c\ 

10 

32700 
-  327° 

45  I 
45  j 

10^650  I 
106650  f 

242730 

* 

169911 

169911 

ci 

— 

—  3270 

34 

80580 

773*0 

" 

54"7 

224028 

^3 

— 

-  3*70 

23 

54510 

51240 

** 

35868 

259896 

^4 

— 

—  ^270 

12 

28440 

25170 

" 

17619 

2775X5 

c* 

— 

—  3270 

i 

2370 

-  900 

-  630 

276885 

v\  —  35*97°  Ibs.  5  v*  —  90,210 ;  vs  =  68,040 ;  z>4  =  47,470  ;  ^5  =  28,500  Ibs. 

48.  Compare  the  relative  amounts  of  iron  required  in  the  webs  of  a 
single-  and  a  double-intersection   Pratt  deck-truss  of  zoo  ft.  span  and 
having  eight  panels.     Panel  live  load  =  L,  panel  dead  load  =  D. 

49.  The  figure  represents  a  pier,  square  in  plan,  supporting  the   ends 
of  two  deck -trusses,  each  200  ft.  long  and  30  ft.  deep.     The  height  of  the 

pier  is  50  ft.  and  is  made  up  of  three  panels,  the 
upper  and  lower  being  each  17  ft.  deep.  Ten 
square  feet  of  bridge  surface  and  ten  square  feet 
of  train  surface  per  lineal  foot  are  subjected  to 
a  wind-pressure  of  40  Ibs.  per  square  foot.  The 
centre  of  pressure  for  the  bridge  is  68  ft.,  and 
for  the  train  86  ft.,  above  the  pier's  base.  The 
wind  also  produces  a  horizontal  pressure  of  4000 
Fia.  450.  Ibs.  at  each  of  the  intermediate  panel  points  on 

the  windward  side  of  the  pier.     Width  of  pier  =  17  ft.  at  top  and  33!  ft. 

at  bottom.     The  bridge  load  =  1600  Ibs.  per  lineal  foot,  live  load  =  3000 

Ibs.  per  lineal  foot.     Determine — 

(a)  The  overturning  moment  (3180  ft.-tons). 

(b)  The  horizontal   force   due  to  the  wind  at  the  top  of  the   pier. 
(61.6  tons.) 

(c)  The  tension  in  the  vertical  anchorage  ties  at  5  and  T.     (Nil.} 

(d)  The  vertical  and  horizontal  reactions  at  T.     (275  and  65.6  tons.) 
Draw  a  diagram  giving  the  wind-stresses  in  all  the  members,  and  in- 
dicate which  are  in  tension  and  which  in  compression. 

Ascertain  whether  the  wind-pressure  of  40  Ibs.  per  square  foot  upon 
a  train  of  empty  cars  weighing  900  Ibs.  per  lineal  foot  will  produce  a 
tension  anywhere  in  the  inclined  posts.  What  will  be  the  tension  in  the 
anchorage  ties  ?  (20.75  tons.) 

Find  the  stresses  in  the  traction  bracing  (i)  when  a  loaded  train  trav- 
elling at  30  miles  an  hour  is  braked  just  as  the  engine  is  over  the  pier 
and  brought  to  rest  in  a  length  of  300  ft.  ;  (2)  when  a  loaded  train  with 
the  engine  over  the  pier  is  started  by  a  sudden  admission  into  the  cylin- 
ders of  steam  at  100  Ibs.  per  square  inch.  Stroke  of  cylinder  =  16  in., 
diameter  of  drivers  =  5  ft. 


EXAMPLES. 


701 


50.  The  figure  represents  one  half  of  one  of  the  piers  of  the  Bouble 
Viaduct.  The  spans  are  crossed  by  two  lattice-gir- 
ders,  14'  9"  deep  and  having  a  deck  platform.  The 
height  of  the  pier  is  183'  9"  and  is  made  up  of  eleven 
panels  of  equal  depth.  Width  of  pier  at  top  =  13' 
ii",  at  bottom  =  67'  7".  With  wind-pressure  at 
55.3  Ibs.  per  square  foot,  the  total  pressure  on  the 
girder,  train,  and  pier  have  been  calculated  to  be  20, 
16.2,  and  20  tons,  acting  at  points  196.2,  210.3,  and 
92.85  ft.,  respectively,  above  the  base.  The  dead 
weight  upon  each  half  pier  is  222^  tons,  of  which  60 
tons  is  weight  of  half  span,  120  tons  the  weight  of 
the  half  pier,  and  42^  tons  the  weight  of  the  train. 
Assuming  that  the  wind-pressure  on  the  pier  is  a 
iiorizontal  force  of  2  tons  at  each  panel  point  on  the 
windward  side,  and  that  the  weight  of  the  pier  may 
be  considered  as  a  weight  of  6  tons  at  each  panel 
point,  determine — 

(a)  The  overturning  moment.  FIG.  451. 

(b)  The  total  horizontal  force  at  the  top  of  the  pier  due  to  the  wind. 

(c)  The  tension  in  each  of  the  vertical  anchorage  ties  at  5  and  T  due 
to  the  wind-pressure. 

(d)  The  vertical  and  horizontal  reactions  at  T. 

Show  that  the  greatest  compressive  stress  occurs  in  the  member  RT, 
and  that  it  amounts  to  422  tons. 

Draw  a  stress  diagram  giving  the  stresses  in  all  the  members,  indi- 
cating which  are  in  tension  and  which  in  compression.  Width  of  pier 
at  A  =  20  ft.,  at  £  =  23$  ft.,  at  C  =  36^  ft. 

What  will  be  the  effect  of  braking  the  train  when  running  at  30  miles 
an  hour,  so  as  to  bring  it  to  rest  within  a  distance  of  220  ft.  ?  Width 
of  pier  in  direction  of  bridge  =  9!  ft.  at  top  and  =  20  ft.  at  bottom. 

Ans. — (a)  9188  ft.-tons;  (ti)  39.9  tons;  (c}  24!  tons,      (d)  Hori- 
zontal reaction  =  59.9  tons  ;  vertical  reaction  =  247  tons. 
The  accompanying  figure  represents  a  portion  of  a  cantilever  truss, 
the  horizontal  distances  of  the  points  A,  Bt  C 
from  the  free  end  being  A  ,  /a ,  I* ,  respectively. 
The  boom  ABC  is  inclined  at  an  angle  a,  and  the 
boom  XYZ  at  an  angle  ft,  to  the  horizon.     Find 
the  deflections  at  the  end  of  the  cantilever  due  to 
X  Y          z    (a)  an  increase  kAB  in  the  length  of  AB\  (2)  an 

FIG.  452.  increase  kiB  Y  in  the  length  of  B  Y ;  (3)  a  decrease 

k*XY  in  the  length  of  XY\  (4)  a  decrease  k±BX  in  the  length  of  BX. 


w 


Ans.— (i) 


BX  sin  ABX 


702 


THEORY  OF  STRUCTURES. 


(3) 


BX 


(4)*« 


BX  cos  a 


sin 


-  /9(cot  BXY  -  cot 


In  the  preceding  question,  if  kl  =  k*  =  k*  =  ki  =  k,  and  if  A  W  is 
parallel  to  ^Jf,  and  AX  to  ^  F,  show  that  the  angle  between  WX  and 
xY  F  after  deformation 

=  2>&(cot  ABX  +  cot  B  YX). 
Hence  also,  if  the  truss  is  of  uniform  depth  d,  show  that  the  "  deviation  " 

2k 

of  the  boom  per  unit  of  length  is  constant  and  equal  to  — • 

d 

52.  Six  bars  have  to  be  arranged  upon  a  steel  pin  ;  each  bar  is  I  in. 
wide  and  is  subjected  to  a  stress  of  64,000  Ibs.  Should  the  bars  be  ar- 


64,000  Ibs. 


FIG.  454.— Method  2. 

ranged  according  to  method  i  or  method  2  ?     Why  ?    Determine  the  di- 
ameter of  the  pin. 

53.  The  accompanying  sketch  represents  one  of  the  pin  connections 
in  a  certain  bridge  which  was  recently  overthrown.  The  two  innermost 
bars  are  web  members  inclined  to  the  horizon  at  an  angle  whose  cosine 

"^^^ 


-[42,100  Ib8. 


FIG.  455. 

is  .815.  The  thickness  of  the  bars  and  the  maximum  stresses  to  which 
they  are  severally  subjected  are  shown  on  the  diagram.  Is  the  3-in. 
wrought-iron  pin  sufficiently  strong? 


CHAPTER   XII. 
SUSPENSION-BRIDGES. 

I.  Cables. — The  modern  suspension-bridge  consists  of  two 
or  more  cables  from  which  the  platform  is  suspended  by  iron 
or  steel  rods.  The  cables  pass  over  lofty  supports  (piers),  and 
are  secured  to  anchorages  upon  which  they  exert  a  direct  pull. 

Chain  or  link  cables  are  the  most  common  in  England  and 
Europe,  and  consist  of  iron  or  steel  links  set  on  edge  and 
pinned  together.  Formerly  the  links  were  made  by  welding 
the  heads  to  a  flat  bar,  but  they  are  now  invariably  rolled  in 
one  piece,  and  the  proportional  dimensions  of  the  head,  which 
in  the  old  bridges  are  very  imperfect,  have  been  much  im- 
proved. 

Hoop-iron  cables  have  been  used  in  a  few  cases,  but  the 
practice  is  now  abandoned,  on  account  of  the  difficulty  attend- 
ing the  manufacture  of  endless  hoop-iron. 

Wire-rope  cables  are  the  most  common  in  America,  and 
form  the  strongest  ties  in  proportion  to  their  weight.  They 
consist  of  a  number  of  parallel  wire  ropes  or  strands,  compactly 
bound  together  in  a  cylindrical  bundle  by  a  wire  wound  round 
the  outside.  There  are  usually  seven  strands,  one  forming  a 
core  round  which  are  placed  the  remaining  six.  It  was  found 
impossible  to  employ  a  seven-strand  cable  in  the  construction 
of  the  East  River  Bridge,  New  York,  as  the  individual  strands 
would  have  been  far  too  bulky  to  manipulate.  The  same  ob- 
jection held  against  a  thirteen-strand  cable  (thirteen  is  the  next 
number  giving  an  approximately  cylindrical  shape),  and  it  was 
finally  decided  to  make  the  cable  with  nineteen  strands.  Seven 
of  these  are  pressed  together  so  as  to  form  a  centre  core,  around 
which  are  placed  the  remaining  twelve,  the  whole  being  con- 
tinuously wrapped  with  wire. 

703 


704 


THEORY   OF  STRUCTURES. 


In  laying  up  a  cable  great  care  is  required  to  distribute  the 
tension  uniformly  amongst  the  wires.  This  may  be  effected 
either  by  giving  each  wire  the  same  deflection  or  by  using 
straight  wire,  i.e.,  wire  which  when  unrolled  upon  the  floor 
from  a  coil  remains  straight  and  shows  no  tendency  to  spring 
back.  The  distribution  of  stress  is  practically  uniform  in  un- 
twisted wire  ropes.  Such  ropes  are  spun  from  the  wires  and 
strands  without  giving  any  twist  to  individual  wires. 

The  back-stay  is  the  portion  of  the  cable  extending  from  an 
anchorage  to  the  nearest  pier. 

The  elevation  of  the  cables  should  be  sufficient  to  allow  for 
settling,  which  chiefly  arises  from  the  deflection  due  to  the  load 
and  from  changes  of  temperature. 

The  cables  may  be  protected  from  atmospheric  influence 
by  giving  them  a  thorough  coating  of  paint,  oil,  or  varnish,  but 
wherever  they  are  subject  to  saline  influence,  zinc  seems  to  be 
the  only  certain  safeguard. 

2.  Anchorage,  Anchorage  Chains,  Saddles. — The  an- 
chorage, or  abutment,  is  a  heavy  mass  of  masonry  or  natural 
rock  to  which  the  end  of  a  cable  is  made  fast,  and  which  re- 
sists by  its  dead  weight  the  pull  upon  the  cable. 


FIG.  456.  FIG.  457.  FIG.  458. 

The  cable  traverses  the  anchorage  as  in  Figs.  456  to  458, 
and  passes  through  a  strong,  heavy  cast-iron  anchor-plate,  and, 
if  made  of  wire  rope,  has  its  end  effectively  secured  by  turning 
it  round  a  dead-eye  and  splicing  it  to  itself.  Much  care,  how- 
ever, is  required  to  prevent  a  wire-rope  cable  from  rusting  on 
account  of  the  great  extent  of  its  surface,  and  it  is  considered 
advisable  that  the  wire  portion  of  the  cable  should  always  ter- 
minate at  the  entrance  to  the  anchorage  and  there  be  attached 
to  a  massive  chain  of  bars,  which  is  continued  to  the  anchor- 
plate  or  plates  and  secured  by  bolts,  wedges,  or  keys. 


ANCHORAGE,  ANCHORAGE    CHAINS,   SADDLES. 

In  order  to  reduce  as  much  as  possible  the  depth  to  which 
it  is  necessary  to  sink  the  anchor-plates,  the  anchor-chains  are 
frequently  curved  as  in  Fig.  458.  This  gives  rise  to  an  oblique 
force,  and  the  masonry  in  the  part  of  the  abutment  subjected 
to  such  force  should  be  laid  with  its  beds  perpendicular  to  the 
line  of  thrust. 

The  anchor-chains  are  made  of  compound  links  consisting 
alternately  of  an  odd  and  an  even  number  of  bars.  The  friction 
of  the  link-heads  on  the  knuckle-plates  considerably  lessens 
the  stress  in  a  chain,  and  it  is  therefore  usual  to  diminish  its 
sectional  area  gradually  from  the  entrance  E  to  the  anchor. 
This  is  effected  in  the  Niagara  Suspension  Bridge  by  varying 
the  section  of  the  bars,  and  in  the  East  River  Bridge  by  vary- 
ing both  the  section  and  the  number  of  the  bars. 

The  necessity  of  preserving  the  anchor-chains  from  rust  is 
of  such  importance  that  many  engineers  consider  it  most 
essential  that  the  passages  and  channels  containing  the  chains 
and  fastenings  should  be  accessible  for  periodical  examination, 
painting,  and  repairs.  This  is  unnecessary  if  the  chains  are 
first  chemically  cleaned  and  then  embedded  in  good  hydraulic 
cement,  as  they  will  thus  be  perfectly  protected  from  all  at- 
mospheric influence. 

The  direction  of  an  anchor-chain  is  changed  by  means  of  a 
saddle  or  knuckle-plate,  which  should  be  capable  of  sliding  to 
an  extent  sufficient  to  allow  for  the  expansion  and  contraction 
of  the  chain.  This  may  be  accomplished  without  the  aid  of 
rollers  by  bedding  the  saddle  upon  a  four-  or  five-inch  thickness 
of  asphalted  felt. 

The  chain,  where  it  passes  over  the  piers,  rests  on  saddles, 
the  object  of  which  is  to  furnish 
bearings  with  easy  vertical  curves. 
Either  the  saddle  may  be  constructed 
as  in  Fig.  459,  so  as  to  allow  the 
cable  to  slip  over  it  with  compara- 
tively little  friction,  or  the  chain  may  be  secured  to  the  saddle, 
and  the  saddle  supported  upon  rollers  which  work  over  a  per- 
fectly true  and  horizontal  bed  formed  by  a  saddle-plate  fixed 
to  the  pier. 


7o6 


THEORY  OF  STRUCTURES. 


3.  Suspenders. — The    suspenders    are  the  vertical  or  in- 
clined  rods  which  carry  the  platform. 


FIG.  460. 


FIG.  461. 


FIG.  462. 


FIG.  463. 


In  Fig.  460  the  suspender  rests  in  the  groove  of  a  cast- 
iron  yoke  which  straddles  the  cable.  Fig.  461  shows  the 
suspender  bolted  to  a  wrought-iron  or  steel  ring  which  em- 
braces the  cable.  When  there  are  more  than  two  cables  in 
the  same  vertical  plane,  various  methods  are  adopted  to  insure 
the  uniform  distribution  of  the  load  amongst  the  set.  In  Fig. 
462,  for  example,  the  suspender  is  fastened  to  the  centre  of  a 
small  wrougnt-iron  lever  PQ,  and  the  ends  of  the  lever  are 
connected  with  the  cables  by  the  equally  strained  rods  PR 
and  QS.  In  the  Chelsea  bridge  the  distribution  is  made  by 
means  of  an  irregularly  shaped  plate  (Fig.  463),  one  angle  of 
which  is  supported  by  a  joint-pin,  while  a  pin  also  passes 
through  another  angle  and  rests  upon  one  of  the  chains. 

The  suspenders  carry  the  ends  of  the  cross-girders  (floor- 
beams),  and  are  spaced  from  5  to  20  ft.  apart.  They  should 
be  provided  with  wrought-iron  screw-boxes  for  purposes  of 
adjustment. 

4.  Curve  of  Cable.  —  CASE  A.  An  arbitrarily  loaded  flexible 
cable  takes  the  shape  of  one  of  the  catenaries,  but  the  true 
catenary  is  the  curve  in  which  a  cable  of  uniform  section  and 
material  hangs  under  its  own  weight  only. 

Let  A  be  the  lowest  point  of  the  cable,  and  take  the  ver- 
tical through  A  as  the  axis  of  y. 

Take  the  horizontal  through  O  as  the  axis  of  x,  the  origin 
O  being  chosen  so  that 

H—mp,     ......     (i) 


/  being  the  weight  of  a  unit  of  length  of  the  cable,  and  H  the 
horizontal  pull  at  A. 


CURVE   OF   CABLE. 


707 


m  or  AO  is  the  parameter,  or  modulus,  of  the  catenary,  and 
OG  is  the  directrix. 

Let  x,  y  be  the  co-ordinates  of  any  point  P,  the  length  of 
the  arc  AP  being  s. 

Draw  the  tangent  PT  and  the  ordinate  PN. 

The  triangle  PNT  is  evidently  a  triangle  of  forces  for  the 
portion  AP,  PN  representing  the  weight  of  AP  (viz.,  ps),  PT 


\ 


O  N 

FIG.  464. 


\ 


the  tangential  pull  T  at  P,  and  NT  the  horizontal  pull  H  at 
A. 

dy  n~AT       PN      PS      s 

•*•  -7-=  tan  PTN=  -=jr =-—=-,       ...     (2) 
dx  TN       H      m 

which  gives  the  differential  equation  to  the  catenary. 
It  may  be  easily  integrated  as  follows : 


ds 


or 


dy^          l~  ~~?         i 

v  I  +  *  =  * 


f.  •  (s) 


ds 


dx 


.•?tog(* 

in 

c  being  a  constant  of  integration. 

When  x  =  o,     s  =  o,     and  therefore     log  m  =  c. 
Hence, 

logM^+^=£ 


708  THEORY  OF  STRUCTURES. 

or 


or 

m 


-  \  /  \ 

-e    -).      ......     (4) 


Again,  ^  ji 

dy       s        i*  * 

-y-  =  -  =  -(e  ™  —  e    »»); 
V 


dx       m       2 
and  hence, 

tn    - 


(5) 


The  constant  of  integration  is  zero,  since  y  =  m  when 
x  —  o. 

The  last  equation  is  the  equation  to  the  catenary,  while  eq. 
(4)  gives  the  length  of  the  arc  A  P. 

By  equations  (4)  and  (5), 


(6) 


Draw  NM  perpendicular  to  PT,  and  let  the  angle  PTN  = 
PNM=e.     Then 

PM  =  PN  sin  0  =  y =  s,    .     .     .     (7) 

<fV  +  m9 
and 

MN  =  PN  cos  0=y— =??==:  m,  .     .    .    (8) 


since  tan  0  =  -f-  =  — . 
or       7« 

Thus,  the  triangle  /W7V  possesses  the  property  that  the 
side  PM  is  equal  to  the  length  of  the  arc  APt  and  the  side 
MN  is  equal  to  the  modulus  w  (=  AO). 

The  area  ^/W6> 

/*x  m*     *  £ 

=y    7</4r  =  — (^«  —  ^"w)  —  ms  =  2  X  triangle  PMN. 


CURVE   OF  CABLE. 
The  radius  of  curvature,  p,  at  P 


709 


y 

m* 


(9) 


PG  being  perpendicular  to  PT. 

At  A,  y  =  m,  and  the  radius  of  curvature  is  also  m.        (id) 


Again, 


Z 

ps 


PT 
_ 


y 


H  —m— 


(II) 

(12) 


pc  being  the  radius  of  curvature  at  A. 

These  catenary  formulae  are  of  little  if  any  use  in  the  design 
and  construction  of  suspension-bridges,  as  they  are  based  upon 
the  assumption  of  a  purely  theoretical  load  which  never  occurs 
in  practice,  viz.,  the  weight  of  a  chain  of  uniform  section  and 
density. 

CASE  B.  Let  the  platform  be  suspended  from  chains  com- 
posed of  a  number  of  links,  and  let  W  be  the  whole  weight  be- 


tween the  lowest  point  O  of  the  chain  and  the  upper  end  Pof 
any  given  link.  Let  the  direction  of  this  link  intersect  that  of 
the  horizontal  pull  (//)  at  O  in  E.  Drop  the  perpendicular  PN. 


THEORY  OF  STRUCTURES. 

The  triangle  PNE  is  evidently  a  triangle  of  forces ;  and  if  the 
angle  PEN  =  6, 

PN       W 


and  hence 

tan  0  oc  W. 

Thus,  by  treating  each  link  separately,  commencing  with  the 
lowest,  the  exact  curve  of  the  chain  may  be  easily  traced. 

Generally  speaking,  the  distribution  of  the  load  may  be 
assumed  to  be  approximately  uniform  per  horizontal  unit  of 
length,  the  load  being  suspended  from  a  number  of  points 
along  each  chain  or  cable  by  means  of  rods.  The  curve  of  the 
cable  will  then  be  a  parabola. 

Let  w  be  the  intensity  of  the  load  per  horizontal  unit  of 
length. 

Let  x,  y  be  the  co-ordinates  of  any  point  P  of  the  cable 
with  respect  to  the  horizontal  6Uf  and  the  vertical  OY  as  axes 
of  x  and  y,  respectively. 

Let  0  be  the  inclination  of  the  tangent  at  P  to  the  horizon- 
tal. The  portion  OP  of  the  cable  is  kept  in  equilibrium  by 
the  horizontal  pull  H  at  O,  by  the  tangential  pull  T  at  P,  and 
by  the  load  wx  upon  OP,  which  acts  vertically  through  the 
middle  point  E  of  ON,  PN  being  the  ordinate  at  P. 

Hence,  the  tangent  at  P  must  also  pass  through  E,  and 
PEN  is  a  triangle  of  forces.  Hence, 

x 

H        2  2H  . 

—  =  -,     or    *•=—  -y,  .....     (i) 
the  equation  to  a  parabola  with  its  vertex  at  O,  its  axis  vertical, 

2H 

and  its  parameter  equal  to . 

Again, 

T  _    PE  _       i 
H~  EN~  cos~0' 


PARAMETER,  ETC.  7 II 

and  hence 

rCOs0  =  ;7=  —  '  (* 

2y  ' 

and  the  horizontal  pull  at  every  point  of  the  cable  is  the  same  as 
that  at  the  lowest  point. 
Also, 

Z£/Jt"2  / 


/  /          x* 

— F  =  W*  \  I    I  +  5« 

xl        V      4/ 


4 
The  radius  of  curvature  at  P 

* 


so  that  the  radius  at  O  is 

H 


or 


5.  Parameter,  etc. — Let  ^,,  ^2  be  the  elevations  of  A 
and  .#,  respectively,  above  the  horizontal  line  COD,  Fig.  465. 
Let  OD  =  alt  OC  =  a,,  and  let  al  +  at=a=  CD. 
By  equation  (i),  Art.  4. 


2ff        a,  a^  a, 


y        i//^a      v^,  +  ^a      y^  +  Vk 

Denote  the  parameter  by  P.     Then 


w 
Also, 


712  THEORY  OF  STRUCTURES. 

If  0,  ,  0a  be  the  values  of  0  at  A  and  B,  respectively. 


tan  6>,  =  2  A-     and     tan  03  =  2  Ai, 


Note.—li  )&,  =  >&,  =  £, 


and  hence 


tan  ft-as=  =  tan  ^,. 


6.  Length  of  Arc  of  Cable.— Let  OP  =  s,  Fig.  465, 


wx 
Since    tan  0  =  -==, 

£2 


or 


seca  OdB  =  -^^r  =  ~</j  cos  ft 


ds  = 

w  cos3 


Hence, 


TT      /»^       7^a  rr 

^  =  ^7Jtan  ^  sec  .0  +  log,  (tan 


cos 


Again, 

tan  6  =  —  x, 
and 


sec  v  = 


WEIGHT   OF   CABLE.  713 

Note.  —  An  approximate  value  of  the  length  of  the  arc  may 
be  obtained  as  follows  : 


/         i  wV\ 
.-.  ds  —  dx\\  +  --  -J7r]>  approximately. 

Integrating  between  O  and  P, 


. 

*  / 

7.  Weight  of  Cable.  —  The  ultimate  tenacity  of  iron  wire 
is  90,000  Ibs.  per  square  inch,  while  that  of  steel  rises  to 
200,000  Ibs.,  and  even  more.  The  strength  and  gauge  of  cable 
wire  may  be  insured  by  specifying  that  the  wire  is  to  have  a 
certain  ultimate  tenacity  and  elastic  limit,  and  that  a  given 
number  of  lineal  feet  of  wire  is  to  weigh  one  pound.  Each  of 
the  wires  for  the  cables  of  the  East  River  Bridge  was  to  have 
an  ultimate  tenacity  of  3400  Ibs.,  an  elastic  limit  of  1600  Ibs., 
and  14  lineal  feet  of  the  wire  were  to  weigh  one  pound.  A  very 
uniform  wire,  having  a  coefficient  of  elasticity  of  29,000,000  Ibs., 
has  been  the  result,  and  the  process  of  straightening  has  raised 
the  ultimate  tenacity  and  elastic  limit  nearly  8  per  cent. 

Let  W,  be  the  weight  of  a  length  a,  (=  OD)  of  a  cable  of 
sufficient  sectional  area  to  bear  safely  the  horizontal  tension  H. 

Let  W^  be  the  weight  of  the  length  s,(  =  OA)  of  the  cable 
of  a  sectional  area  sufficient  to  bear  safely  the  tension  7\  at  A. 

Let  /be  the  safe  inch-stress. 

Let  q  be  the  specific  weight  of  the  cable  material. 
Then 


and 


714  THEORY  OF  STRUCTURES. 


••«'. -»-.| -<-*+!•••<- +.4 


or 


+ V),  nearly. 

A  saving  may  be  effected  by  proportioning  any  given  section 
to  the  pull  across  that  section. 

At  any  point  (x,  y)  the  pull  =  H  sec  0,  and  the  correspond- 

ing  sectional  area  — j — .     The  weight  per  unit  of  length 

=  — -7. — g,  and  the  total  weight  of  the  length  sl  (=  OA)  is 
r  sec  0    ds 


Hq 
f 


Hence, 


and  also 


The  weight  of  a  cubic  inch  of  steel  averages  .283  Ib. 

The  weight  of  a  cubic  inch  of  wrought-iron  averages  .278  Ib. 

IT 

The  volume  in  inches  of  the  cable  of  weight  W1=  \2al—p  . 


DEFLECTION  OF  CABLE.  715 

W, 

--     =  .283  lb.  or  .278  lb., 


according  as  the  cable  is  made  of  steel  or  iron. 

Let  the  safe  inch-stress  of  steel  wire  be  taken  at  33,960  Ibs., 
of  the  best  cable-iron  at  14,958  Ibs.,  and  of  the  best  chain-links 
at  9972  Ibs.  Then 


W,  =  Ha,  X  .283  X  =  for  steel  cables  ; 


Wt  =  Ha,  X  .278  X  -~-g  =  for  iron  cables  ; 


t  =  Ha,  X  .278  X  ~    =  for  link  cables. 


Note.  —  About  one-eighth  may  be  added  to  the  net  weight  of 
a  chain-cable  for  eyes  and  fastenings. 

8.  Deflection  of  a  Cable  due  to  an  Elementary  Change 
in  its  Length. 

By  the  corollary  of  Art.  6  the  total  length  (S)  of  the  cable 
A  OB  is 


Now  a,  and  #,  are  constant  ;  h^  —  h9  is  also  constant,  and 
therefore  dh^  =  dhv     Hence, 


If  the  alteration  in  length  is  due  to  a  change  of  /°  in  the 
temperature, 

dS  =  etS, 

e  being  the  coefficient  of  linear  expansion  and  =  ^ ^~ 

per  degree  Fahr.  for  wrought-iron.  • 


THEORY  OF  STRUCTURES. 

In  England  the  effective  range  of  temperature  is  about  60° 
Fahr.,  while  in  other  countries  it  is  usual  to  provide  for  a  range 
of  from  100°  to  150°  F. 

If  the  alteration  is  due.  to  a  pull  of  intensity /per  unit  of 
area, 

dS  =  gS, 

E  being  the  coefficient  of  elasticity  of  the  cable  material. 

If  h,  =  h,  =  h, 

a  16  h 

# ,  =  at  =  —  ,     and     dS  = dh. 

9.  Curve  of  Cable  from  which  the  load  is  suspended  by  a 
series  of  sloping  rods. 


r      o       E      N 

FIG.  466. 

Let  O  be  the  lowest  point  of  such  a  cable.  Let  the  tangent 
at  O,  and  a  line  through  O  parallel  to  the  suspenders,  be  the 
axes  of  x  and  y,  respectively. 

Let  w'  be  the  intensity  of  the  oblique  load.  Consider  a 
portion  OP  of  the  cable,  and  let  the  co-ordinates  of  P  with 
respect  to  OX,  OY  be  x  and  y. 

Draw  the  ordinate  PN,  and  let  the  tangent  at  P  meet  ON 
in  E. 

As  before,  PNE  is  a  triangle  of  forces,  and  E  is  the  middle 
point  of  ON.  Then 

PN      2 


_         _    ^  2  _ 

~~'  ~~  ~y' 


the  equation  to  a  parabola  with  its  axis  parallel  to  OY  and  its 
focus  at  a  point  S,  where  ^SO  —  —  ,-  . 


CURVE   OF   CABLE   WITH  OBLIQUE   SUSPENDERS. 

Cor.  i.  Let  the  axis  meet  the  tangent  at  O  in  Tf,  and  let 
its  inclination  to  OX  be  i. 

Let  A  be  the  vertex,  and  ON'  a  perpendicular  to  the  axis. 

Then 

SO  =  ST'  =  SA  +  A  Tf  =  SA  +  AN'. 
But       AS  .AN'  =  ON"  =  NfT'*tan*i 


.-.  AS  =  AN'  tan2  i,     and     SO  =  AS(i  +  cot2  *)  ='        -.. 

smaz 

Hence,  the  parameter  —  4^45  =  ^SO  sin2  /. 

Cor.  2.  Let  P  be  the  oblique  load  upon  the  cable  between 
O  and  P. 

Let  Q  be  the  total  thrust  upon  the  platform  at  E. 

11     w  "     "    load  per  horizontal  unit  of  length. 

"     q    "     "    rate  of  increase  of  thrust  along  platform. 

"     /     "     "    length  of  PE. 
Then 

w 
w  —  —  —  .  ,     and     q  =  w  cot  z  ; 


sm  i  sin  i 


p  =  ff       ^ 


x         y 
x* 

t*  —  y  +  —  +  *y  cos  i. 

4 

Cor.  3.  Let  s  be  the  length  of  OP,  and  let  6  be  the  inclina- 
tion of  PE  to  OY.     Then 

s  =  AP-  AO 

tan  (9°°  -  ^  sec(9°°  -  ^ 

+  log,jtan  (90°  —  &)  +  sec  (90°  -  0)}  -  tan  (90°-  *)  sec  (go0—*) 

-  log,  {  tan  (90°  -  1  )  +  sec  (90°  -  i)  }  j 

//"sin*  «  (  ;-•.  cot  #  +  cosec  <5»  ) 

—  ---  :  —  -J  cot  0  cosec  6  —  cot  /  cosec  i  4-  log,  -  r^ 

2W       (  cot  z  -f~  cosec  z  ) 


7i8 


THEORY  OF  STRUCTURES. 


It  may  be  easily  shown,  as  in  the  Note  to  Art.  6,  that  ap- 
proximately 


.  . 
s  =  x  4-  y  cos  t  -4- 


sin2  * 


t  .  . 

X  -j-J/COSZ 


10.  Pressure  upon  Piers,  etc. 

Let  71,  be  the  tension  in  the  main  cable  at  A. 
"     T;  "     "          "        "     "    back-stay  at  A 
"     «?,  /?  be  the  inclinations  to  the  horizontal  of  the  tangents 
at  A  to  the  main  cable  and  back-stay,  respectively. 

The  total  vertical  pressure  upon  the  pier  at  A 

—  71,  sin  a  -j-  7",  sin  y#  —  /?. 
The  total  resultant  horizontal  force  at  ^4 
=  7i  cos  «f  ~  Tzcos  fi  =  Q. 

If  the  cable  is  secured  to  a  saddle  which  is  free  to  move 
horizontally  on  the  top  of  the  pier  (Fig.  467), 

Q  —  the  frictional  resistance  to  the  tendency  to  motion, 

or       Q  =  ^Rt 
fa  being  the  corresponding  coefficient  of  friction. 


FIG.  467. 


Let  A  Fig.  468,  be  the  total  height  of  the  pier,  and  let  W 
its  weight. 

Let  FG  be  the  base  o 
of  the  centre  of  pressure. 


4 

be  its  weight. 

Let  FG  be  the  base  of  the  pier,  and  K  the  limiting  position 

,  *  f 


AUXILIARY  OR   STIFFENING    TRUSS. 

Let  /,  q  be  the  distance  of  P  and  W,  respectively,  from 
Then 

pp  _L  Wq 
for  stability  of  position  Q  =  --  />""      » 

and  for  stability  of  friction,  when  the  pier  is  of  masonry, 


.     the  coefficient  of  friction  of  the  masonry. 


If  /^  is  sufficiently  small  to  be  disregarded,  Q  is  approxi- 
mately nil,  and  7,  cos  a  =  7",  cos  fi  =  H.  The  pressure  upon 
the  pier  is  now  wholly  vertical  and  is  —  //"(tan  a  -j-  tan  /?). 

When  the  cable  slides  over  smooth  rounded  saddles  (Fig. 
459),  the  tensions  7,  and  7,  are  approximately  the  same. 

Thus, 

R  =  T,  (sin  a  +  sin  ft)     and     g  =  Tt(cos  a  —  cos  ft). 

It  a  =  fi,  Q  =  o,  and  the  pressure  upon  the  pier  is  wholly 
vertical,  its  amount  being  27,  sin  a. 

The  piers  are  made  of  timber,  iron,  steel,  or  masonry,  and 
allow  of  great  scope  in  architectural  design. 

The  cable  should  in  no  case  be  rigidly  attached  to  the  pier, 
unless  the  lower  end  of  the  latter  is  free  to  revolve  through  a 
small  angle  about  a  horizontal  axis. 

II.  Auxiliary  or  Stiffening  Truss.  —  The  object  of  a  stiff- 
ening truss  (Fig.  469)  is  to  distribute  a  passing  load  over  the 
cable  in  such  a  manner  that  it  cannot  be  distorted.  The  pull1 
upon  each  suspender  must  therefore  be  the  same,  and  this  vir- 


tually assumes  that  the  effect  of  the  extensibility  of  the  cable 
and  suspenders  upon  the  figure  of  the  stiffening  truss  may  be 
disregarded. 


720  THEORY  OF  STRUCTURES. 

The  ends  O  and  A  must  be  anchored,  or  held  down  by 
pins,  but  should  be  free  to  move  horizontally. 

Let  there  be  n  suspenders  dividing  the  span  into  (n  -\-  i) 
equal  segments  of  length  a. 

Let  P  be  the  total  weight  transmitted  to  the  cable,  and  z 
the  distance  of  its  centre  of  gravity  from  the  vertical  through  O. 

Let  T  be  the  pull  upon  each  suspender. 

Taking  moments  about  O, 


/  being  the  length  of  OA. 

Also,  if  t  is  the  intensity  of  pull  per  unit  of  span, 

r 

tl  =  nT,     and  hence     Pz  —  t  —  . 

Let  there  be  a  central  suspender  of  length  s.     There  will, 
therefore,  be  -  suspenders  on  each  side  of  the  centre. 

r 

The  parameter  of  the  parabola  =  —=-  . 
Hence,  the  total  length  of  all  the  suspenders 


If  there  is  no  central  suspender,  i.e.,  if  n  is  even, 
the  total  length  =  (n  —  i)(s  +  -"•  -j—J. 
Denote  the  total  length  of  suspenders  by  L.     Then 
the  stress-length  =  TL  =  ~PL. 


AUXILIARY   OR   STIFFENING    TRUSS.  ? '21 

Let  w  be  the  uniform  intensity  of  the  dead  load. 

CASE  I.    The  bridge  partially  loaded. 

Let  w'  be  the  maximum  uniform  intensity  of  the  live  load, 
and  let  this  load  advance  from  A  and  cover  a  length  AB. 

Let  OB  —  x,  and  let  Rl ,  R^  be  the  pressures  at  O  and  A, 
respectively. 

For  equilibrium, 

R^R^  +  tl-wl-  «/(/ -  *)  =  o  ;  •     •     •     (i) 
J_z/-^(/-*)°       =0.    ...     (2) 


Also,  since  the  whole  of  the  weight  is  to  be  transmitted 
through  the  suspenders, 


>(l  —  x)  .......     (3) 

From  eqs.  (i),  (2),  and  (3), 


(4) 


which  shows  that  the  reactions  at  O  and  A  are  equal  in  mag- 
nitude but  opposite  in  kind.     They  are  evidently  greatest  when 

x  •=  -,  i.e.,  when  the  live  load  covers  half  the  bridge,  and  the 

w'l 
common  value  is  then  —r-  . 

O 

The  shearing  force  at  any  point  between  O  and  B  distant  x' 
from  O 


which    becomes   —  j(l  —  x)  —  —  Rl  =  R^  when   x'  equal   x. 

Thus  the  shear  at  the  head  of  the  live  load  is  equal  in  magni- 
tude to  the  reaction  at  each  end,  and  is  an  absolute  maximum 


722  THEORY  OF  STRUCTURES. 

when  the  live  load  covers  half  the  bridge.     The  web  of  the 

rf 

8 


truss  must  therefore  be  designed  to  bear  a  shear  of  —^-  at  the 


centre  and  ends. 

Again,  the  bending  moment  at  any  point  between  O  and  B 
distant  x'  from  O 

=  Rlx'  +  tx"  =  ~l-=(x»-xx'),     .     .     (6) 


which  is  greatest   when  *'  =-,  i.e.,  at  the  centre  of  OB,  its 

VOJ       /    _     JK 

value  then  being  —  —  —  f~x*-     Thus,  the  bending  moment  is 

O  / 

d  *  2 

an  absolute  maximum  when  ~j~(tx*  —  **)  =  o,  i.e.,  when  x  •=.  -/,. 

w' 
and  its  value  is  then  ---  /*. 

54 

The  bending  moment  at  any  point  between  B  and  A  dis- 
tant x'  from  O 


"  -  -  x?  =       j(x>  -  x)(l-  x'\  (7) 


d 
which  is  greatest  when  ~r~t{(^'  ~~  *Yt  ~  x'}\  =  °»  i-e->  when 

14-x 
x'  = ,   or   at    the    centre   of  AB,  its  value    then    being 

w'  x 


—  —(I  —  x}*.     Thus,  the  bending  moment  is  an  absolute  niaxi- 

0      / 

3 


mum  when  -j-\x(l  —  x)*\  =  o,  i.e.,  when  x  =  — ,  and  its  value  is 


then  +       /'. 
54 

Hence,  the  maximum  bending  moments  of  the  unloaded  and 
loaded  divisions  of  the  truss  are  equal  in  magnitude  but  opposite 
in  direction,  and  occur  at  the  points  of  triscction  (D,  C)  of  OA 


AUXILIARY  OR    STIFFENING    TRUSS.  723 

when  the  live  load  covers  one-third  (AC)  and  two-thirds  (AD)  of 
the  bridge,  respectively. 

Each  chord  must  evidently  be  designed  to  resist  both 
tension  and  compression,  and  in  order  to  avoid  unnecessary 
nicety  of  calculation,  the  section  of  the  truss  may  be  kept  uni- 
form throughout  the  middle  half  of  its  length. 

CASE  II.  A  single  concentrated  load  W  at  any  point  B  of 
the  truss.  W  now  takes  the  place  of  the  live  load  of  intensity 
<w'. 

The  remainder  of  the  notation  and  the  method  of  pro- 
cedure being  precisely  the  same  as  before,  the  corresponding 
equations  are 

R1  +  R,  +  (t-w)l-W  =  o (i') 

'-W(l-x)  =  o (2') 


2 

W 


(3') 


(4') 


which  shows  that  the  reactions  at  O  and  A  are  equal  in  mag- 
nitude but  opposite  in  kind.  They  are  greatest  when  x  =  o 
and  when  x  =  /,  i.e.,  when  W  is  either  at  O  or  at  A,  and  the 

W 
common  value  is  then  —  . 

2 

The  shearing  force  at  any  point  between  O  and  B  distant  x' 
from  O 

W 


W 
which  is  a  maximum  when  x'  =  x,  and  its  value  is  then  —  . 

2 

The  web   must  therefore   be  designed  to  bear  a  shear  of 

W 

—  throughout  the  whole  length  of  the  truss. 


724  THEORY  OF  STRUCTURES. 

Again,  the  bending  moment  at  any  point  between  O  and 
B  distant  x'  from  O 


First y  let  w  <  — .     The  bending  moment  is  positive  and  is  a 
maximum  when  x'  =.  x,  its  value  then  being 

+  —(/ 
21 

Next,\&.  x  >— .     The  bending  moment  is  then  negative  va^ 

is  a  maximum  when  x'  =  x  —  -,  its  value  then  being 

Wi 


The  bending  moment  at  any  point  between  B  and  A  dis 
tant  x'  from  O 


=  R,x'  +  (t-w)--  W(X'  -x)  =  -j(x'  -  /)-  - 
which  is  a  maximum  when 


I  W  I         l\ 

i.e.,  when  x'  =  x  -4-  -,  and  its  value  is  then  —  —  -Ax  —  —  }  . 

1     2  2/\  21 

Note.  —  The  stiffening  truss  is  most  effective  in  its  action,  but 
adds  considerably  to  the  weight  and  cost  of  the  whole  struc- 
ture. Provision  has  to  be  made  both  for  the  extra  truss  and 
for  the  extra  material  required  in  the  cable  to  carry  this  extra 
load. 


AUXILIARY  OR   STIFFENING    TRUSS.  725 

Stiffening  Truss  hinged  at  the  Centre. — Provision  may  be 
made  for  counteracting  the  straining  due  to  changes  of  tem- 
perature by  hinging  the  truss  at  the  centre  E. 

Let  a  live  load  of  intensity  w  advance  from  A. 

First,  let  the  live  load  cover  a  length  AB  =  x  f  >  — ). 

Let  R^ ,  R^  be  the  pressures  at  O,  A,  respectively. 
The  equations  of  equilibrium  are 

£l  +  Rt  +  (t-iv)/-iv'x  =  0;    ....     (I) 

--*-'="    •    •    •    W 


(3) 


Eqs.  (2)  and  (3)  being  obtained  by  taking  moments  about  E. 
Hence, 

w' 
t-w=-  -(/3  -  4**  +  2*')  ;       ...     (4) 


;  •    •    •    •    (5) 


Next,  iet  the  live  load  cover  the  length  BO  (  <  -)• 


Let  AB  =  x  as  before,  and  let  JR/t  RJ,  t'  be  the  new  values 
of  Rlt  R^  t,  respectively. 

The  equations  of  equilibrium  are  now 

«'X-«'V-~*)=o;      .    .    (7) 


726  THEORY  OF  STRUCTURES. 

*,'-  +  (;'-Hg=o; (9) 

and  hence, 

/'-W  =  2-(/-  *)'[=-(/-«,  -a-')]'  (10) 


I  W 

\i -*•)>(= -R) (12) 


Diagram  of  Maximum  Shearing  Force.  —  The  shear  at  any 
point  distant  z  from  A  in  the  unloaded  portion  BO  when  the 
live  load  covers  AB 

^R^(t-w)(l-z)  .........     (13) 

>-«,  _«/)(/_*)} 
'  -  «/)(/-  z)  -  w\l-  z) 


=  minus  the  shear  at  the  same  point  when  AB  is 
unloaded  and  the  live  load  covers  BO. 

For  a   given  value  of   z  the  maximum  shear,  positive   or 
negative,  at  any  point  of  OB,  is  found  by  making  (see  eq.  (13)  ) 


or 


3*)  -        (/-*K-  4/+  4*)  =  O, 


or 

74#  —  2/ 
jr=z/T-      ••••••    (H) 


AUXILIARY   OR   STIFFENING    TRUSS. 

Hence,  by  eqs.  (4),  (5),  (13),  (14), 


727 


I  —  2X 

the  maximum  shear  =  ±  %w'x— ,    .    .     (15) 


/-* 


and   may  be  represented  by  the  ordinate      m 
{positive  or  negative)  of  the  curve  mnpq. 
For  example,  at  the  points  defined  by 


*=     /,        v,     */, 

the  shears  are  greatest  when 

*=     K          |/,       i/, 

and  their  values  are,  respectively, 

o. 


o 

VI 


n     I 


FIG.  470. 


Again,  the  shear  at  any  point   distant   z  from   A  in  the 
loaded  portion  BE  when  the  live  load  covers  AB 

=  ^  +  (/  _  «,)(/-  ^)  -  W'(^  -  ^)  ....     (16) 


-  w  —- 


w\l  —  x) 


=  minus  the  shear  at  the  same  point  when  AB  is 
unloaded  and  the  live  load  covers  BO. 


Hence,  by  eqs.  (4),  (5),  (16), 


\w' 


the  shear  =  q=  -  -^(l  -  ^z](l  -  x)\     .     (17) 

increasing  for  a  given  value  of  z  with  /  —  x,  and,  therefore,  a 
maximum  when  x  =  z.     Thus, 

the  maximum  shear  =  =f  ~r(^  "~  4Jtr)(^~-  ^t    •     (J8) 


and  occurs  immediately  /«  front  of  the  load  when  it  covers 
AB,  and  immediately  behind  the  load  when  it  covers  BO.     It 


728  THEORY  OF  STRUCTURES. 

may  be  represented  by  the  ordinate  (positive  or  negative]  of 
the  curve  orsq. 

For  example,  at  the  points  defined  by 


the  maximum  shears  given  by  eq.  (18)  are,  respectively, 
o,     ±yfrw7,     ±TVe>'/,     ±AV*//,     ±\w'l. 


Diagram  of  Maximum  Bending  Moment.  —  The  bending 
moment  at  any  point  in  BO  distant  z  from  A  when  the  live 
load  covers  AB 


(19) 


*} 


=  minus  the  bending  moment  at  the  same  point  when 

the  live  load  covers  BO. 
Hence,  by  eqs.  (4),  (5),  (19),  the  bending  moment 

I  w'  I  w' 

=  ±-2-j(f  -  At*  +  3**X/  -*)  =F  -  -f(l*  -  Al*  +  z**)(l  -*)1. 

For  a  given  value  of  z  this  is  a  maximum  and  equal  to 

W!  zl  —  zl  —  2.Z  2lz 

±T    (1-2,)        when     *  = 


Thus,  the  maximum  bending  moment  may  be  represented 
by  the  ordinate  (positive  or  negative]  of  a  curve. 
For  example,  at  the  points  defined  by 

*  =  /,         v,         */,  t/,       '-, 


AUXILIARY  OR    STIFFENING    TRUSS. 
the  bending  moments  are  greatest  when  ^  — 

v,      Ty,         v,          v,       V-, 

their  values  being,  respectively, 


The  absolute  maximum  bending  moment  may  be  found  as 
follows  : 

For  a  given  value  of  x  the  bending  moment  (see  eq.  (19))  is 
a  maximum  when 


or 


/  —  ze; 
Hence,  the  maximum  bending  moment 


h  "      - 


2  *  —  w  "      -   8     /'  —  4/*r  -f  2*2  ' 

It  will  be  an  absolute  maximum  for  a  value  of  x  found  by  put- 
ting its  differential  with  respect  to  x  equal  to  nil. 
This  differential  easily  reduces  to 


x  =  f/  is  an  approximate  solution  of  this  equation,  and  the  cor- 
responding maximum  bending  moment  =  -^-^w'T. 

The  preceding  calculations  show  that  at  every  point  in  its 
length  the  truss  may  be  subjected  to  equal  maximum  shears  and 
equal  maximum  bending  moments  of  opposite  signs. 

Again,  it  may  be  easily  shown,  in  a  similar  manner,  that 


73°  THEORY   OF   STRUCTURES. 

when  a  single  weight  W  travels  over  the  truss, 

the  maximum  positive  shear  at  a  distance  z  from  A 

W 
=  7r(2/'  -5/2 

the  maximum  negative  shear 

W 

either  =  -(/2  -  5/*+  42*) 


I   W 
or  =  -  -j 


and  the  maximum  bending  moment 
W 

=    ±T*(/  -*)(/-  2,). 


12.  Suspension-bridge  Loads.  —  The  heaviest  distributed 
load  to  which  a  highway  bridge  may  be  subjected  is  that  due 
to  a  dense  crowd  of  people,  and  is  fixed  by  modern  French 
practice  at  82  Ibs.  per  square  foot.  Probably,  however,  it  is 
unsafe  to  estimate.  the  load  at  less  than  from  100  to  140  Ibs.  per 
square  foot,  while  allowance  has  also  to  be  made  for  the  con- 
centration upon  a  single  wheel  of  as  much  as  36,000  Ibs.,  and 
perhaps  more. 

A  moderate  force  repeatedly  applied  will,  if  the  interval 
between  the  blows  corresponds  to  the  vibration  interval  of  the 
chain,  rapidly  produce  an  excessive  oscillation  (Chap.  Ill, 
Cor.  2,  Art.  24).  Thus,  a  procession  marching  in  step  across 
a  suspension-bridge  may  strain  it  far  more  intensely  than  a 
dead  load,  and  will  set  up  a  synchronous  vibration  which  may 
prove  absolutely  dangerous.  For  a  like  reason  the  wind 
usually  sets  up  a  wave-motion  from  end  to  end  of  a  bridge. 

The  factor  of  safety  for  the  dead  load  of  a  suspension-bridge 
should  not  be  less  than  2^  or  3,  and  for  the  live  load  it  is 
advisable  to  make  it  6.  With  respect  to  this  point  it  may  be 
remarked  that  the  efficiency  of  a  cable  does  not  depend  so 
much  upon  its  ultimate  strength  as  upon  its  limit  of  elasticity, 


MODIFICATIONS   OF   THE   SIMPLE    SUSPENSION-BRIDGE.    731 

and  so  long  as  the  latter  is  not  exceeded  the  cable  remains  un- 
injured. For  example,  the  breaking  weight  of  one  of  the  1 5-inch 
cables  of  the  East  River  Bridge  is  estimated  to  be  12,000  tons, 
its  limit  of  elasticity  being  81 18  tons ;  so  that  with  i£  only  as  a 
factor  of  safety,  the  stress  would  still  fall  below  the  elastic 
limit  and  have  no  injurious  effect.  The  continual  application 
of  such  a  load  would  doubtless  ultimately  lead  to  the  destruc- 
tion of  the  bridge. 

The  dip  of  the  cable  of  a  suspension-bridge  usually  varies 
from  ^  to  Y1^  of  the  span,  and  is  rarely  as  much  as  y1^,  except 
for  small  spans.  Although  a  greater  ratio  of  dip  to  span  would 
give  increased  economy  and  an  increased  limiting  span,  the 
passage  of  a  live  load  would  be  accompanied  by  a  greater  dis- 
tortion of  the  chains  and  a  larger  oscillatory  movement. 
Steadiness  is  therefore  secured  at  the  cost  of  economy  by 
adopting  a  comparatively  flat  curve  for  the  chains. 

13.  Modifications  of  the  Simple  Suspension-bridge. — 
The  disadvantages  connected  with  suspension-bridges  are  very 
great.  The  position  of  the  platform  is  restricted,  massive 
anchorages  and  piers  are  generally  required,  and  any  change  in 
the  distribution  of  the  load  produces  a  sensible  deformation  in 
the  structure.  Owing  to  the  want  of  rigidity,  a  considerable 
vertical  and  horizontal  oscillatory  motion  may  be  caused,  and 
many  efforts  have  been  made  to  modify  the  bridge  in  such  a 
manner  as  to  neutralize  the  tendency  to  oscillation. 

(a)  The  simplest  improvement  is  that  shown  in  Fig.  472, 
where  the  point  of  the  cable  most  liable  to  deformation  is 
attached  to  the  piers  by  short  straight  chains  AB. 


FIG.  472. 


(b)  A  series  of  inclined  stays,  or  iron  ropes,  radiating  from 
the  pier-saddles,  may  be  made  to  support  the  platform  at  a 
number  of  equidistant  points  (Fig.  473).  Such  ropes  were  used 
in  the  Niagara  Bridge,  and  still  more  recently  in  the  East  River 


73  2  THEORY  OF   Sl^RUCTURES. 

Bridge.  The  lower  ends  of  the  ropes  are  generally  made  fast 
to  the  top  or  bottom  chord  of  the  bridge-truss,  so  that  the  cor- 
responding chord  stress  is  increased  and  the  neutral  axis  pro- 
portionately displaced.  To  remedy  this,  it  has  been  proposed 
to  connect  the  ropes  with  a  horizontal  tie  coincident  in  position 
with  the  neutral  axis.  Again,  the  cables  of  the  Niagara  and 


FIG.  473. 

East  River  bridges  do  not  hang  in  vertical  planes,  but  are  in- 
clined inwards,  the  distance  between  them  being  greatest  at 
the  piers  and  least  at  the  centre  of  the  span.  This  drawing  in 
adds  greatly  to  the  lateral  stability,  which  may  be  still  further 
increased  by  a  series  of  horizontal  ties. 

(c)  In  Fig.  474  two  cables  in  the  same  vertical  plane  are 
diagonally  braced  together.    In  principle  this  method  is  similar 


FIG.  474. 

to  that  adopted  in  the  stiff ening  truss  (discussed  in  Art.  1 1),  but 
is  probably  less  efficient  on  account  of  the  flexible  character  of 
the  cables,  although  a  slight  economy  of  material  might  doubt- 
less be  realized.  The  braces  act  both  as  struts  and  ties,  and 
the  stresses  to  which  they  are  subjected  may  be  easily  calcu- 
lated. 

(d)  In  Fig.  475  a  single  chain  is  diagonally  braced  to  the 
platform.    The  weight  of  the  bridge  must  be  sufficient  to  insure 


FIG.  475. 

that  no  suspender  will  be  subjected  to  a  thrust,  or  the  efficiency 
of  the  arrangement  is  destroyed.     An  objection  to  this  as  well 


MODIFICATIONS  OF   THE   SIMPLE   SUSPENSION-BRIDGE.    733 


as  to  the  preceding  method  is  that  the  variation  in  the  curva- 
ture of  the  chain  under  changes  of  temperature  tends  to  loosen 
and  strain  the  joints. 

The  principle  has  been  adopted  (Fig.  476)  with  greater  per- 
fection in  the  construction  of  a  foot-bridge  at  Frankfort.     The 


FIG.  476. 


girder  is  cut  at  the  centre,  the  chain  is  hinged,  and  the  rigidity 
is  obtained  by  means  of  vertical  and  inclined  braces  which  act 
both  as  struts  and  ties. 

(e)  In  Fig.  477  the  girder  is  supported  at  several  points  by 


FIG.  477. 

straight  chains  running  directly  to  the  pier-saddles,  and  the 
chains  are  kept  in  place  by  being  hung  from  a  curved  chain  by 
vertical  rods. 

(/)  It  has  been  proposed  to  employ  a  stiff  inverted  arched 
rib  of  wrought-iron  instead  of  the  flexible  cable.  All  straining 
action  may  be  eliminated  by  hinging  the  rib  at  the  centre  and 
piers,  and  the  theory  of  the  stresses  developed  in  this  tension 
rib  is  precisely  similar  to  that  of  the  arched  rib,  except  that 
the  stresses  are  reversed  in  kind. 

(g)  The  platform  of  every  suspension-bridge  should  be 
braced  horizontally.  The  floor-beams  are  sometimes  laid  on 
the  skew  in  order  that  the  two  ends  of  a  beam  may  be  sus- 
pended from  points  which  do  not  oscillate  concordantly,  and 
also  to  distribute  the  load  over  a  greater  length  of  cable. 


734  THEORY  OF  STRUCTURES. 


EXAMPLES. 

1.  The  span  of  a  suspension-bridge  is  200  ft.,  the  dip  of  the  chains  is 
80  ft.,  and  the  weight  of  the  roadway  is  i  ton  per  foot  run.    Find  the  ten- 
sions at  the  middle  and  ends  of  each  chain.  Ans.  31  \  tons ;  58.94  tons. 

2.  Assuming  that  a  steel  rope  (or  a  single  wire)  will  bear  a  tension  of 
15  tons  per  square  inch,  show  that  it  will  safely  bear  its  own  weight  over 
a  span  of  about  one  mile,  the  dip  being  one-fourteenth  of  the  span. 

Ans.  Max.  tension  =  33,074  Ibs. 

3.  Show  that  a  steel  rope  of  the  best  quality,  with  a  dip  of  one-seventh 
of  the  span,  will  not  break  until  the  span  exceeds  7  miles,  the  ultimate 
strength  of  the  rope  being  60  tons  per  square  inch. 

Ans.  Max.  tension  =  59.545  tons  per  square  inch. 

4.  The  river  span  of  a  suspension-bridge  is  930  ft.  and  weighs  5976 
tons,  of  which  1439  tons  are  borne  by  stays  radiating  from  the  summit 
of  each  pier,  while  the  remaining  weight  is  distributed   between  four 
I5~in.  steel-wire  cables,  producing  in  each  at  the  piers  a  tension  of  2064 
tons.     Find  the  dip  of  the  cables.  Ans.  66.44  ft. 

The   estimated  maximum  traffic  upon  the   river  span  is  1311   tons 
uniformly  distributed.     Determine  the  increased  stress  in  the  cables. 

Ans.  596.4  tons. 

To  what   extent  might  the  traffic  be  safely  increased,  the  limit  of 
elasticity  of  a  cable  being  8116  tons,  and  its  breaking  stress  12,300  tons  ? 

Ans.  To  13,303  tons  uniformly  distributed. 

5.  If  the  span  =  /,  the  total   uniform  load  =  W,  and  the  dip  =  — , 

show  that  the  maximum  tension  =  1.58  W,  the  minimum  tension 
=  1.5  IV,  the  length  of  the  chain  =  i.oiS/,  and  find  the  increase  of  dir> 
corresponding  to  an  elongation  of  i  in.  in  the  chain. 

6.  A  cable  weighing  p  Ibs.  per  lineal  foot  of  lengthjs  stretched  be- 
tween supports  in  the  same  horizontal  line  and  20  ft.  apart.     If  the  max- 
imum deflection  is  £ft.,  determine  the  greatest  and  least  tensions. 

Ans.  Parameter  m  =  100  ft.;  max.  tension  =  ioo£/;    min.  ten- 
sion =  ioo^. 

7.  A   light  suspension-bridge  carries  a  foot-path  8   ft.  wide  over  a 
river  90  ft.  wide  by  means  of  eight  equidistant  suspending  rods,  the  dip 
being  10  ft.    Each  cable  consists  of  nine  straight  links.     Find  their  several 
lengths.     If  the  load  upon  the  platform  is  120  Ibs.  per  square  foot,  and 


EXAMPLES.  735 

if  one-fourth  of  the  load  is  borne  by  the  piers,  find  the  sectional  areas 
of  the  several  links,  allowing  10,000  Ibs.  per  square  inch. 

Ans.  Lengths  in  ft.,  10 ;    10.049;  10.198;  10.44;  10.77. 

Tensions  in  Ibs.,  45000;  45004/101  ;  45004/104;  4500^109; 

4500  4/1 1 6. 
Areas  in  sq.  in.,  4.5  ;  4.522  ;  4.59 ;  4.698  ;  4.847. 

8.  A  suspension-bridge  of   200  ft.  span  and  20   ft.  dip   has  4^  sus- 
penders on  each  side ;  the  dead  weight  =  3000  Ibs.  per  lineal   foot ;  the 
live  load  =  2000  Ibs.  per  lineal  foot.     Find  the  maximum  pull  on  a  sus- 
pender, the  maximum  bending  moment  and  the  maximum  shear  on  the 
stiffening  truss.  Also,  findthe  elongation  in  the  chain  due  to  the  live  load. 

Ans.  Max.  pull  =  12,500  Ibs. ;  max.  shear  =  30,000  Ibs.;  max. 
B.M.  =  1,066,666$  ft. -Ibs. ;  elongation  =  89,600,000  -f-  EA,  A 
being  sectional  area  of  a  cable,  and  E  the  coefficient  of  elas- 
ticity. 

9.  A  foot-path  8  ft.  wide  is  to  be  carried  over  a  river  100  ft.  wide  by 
two  cables  of  uniform  sectional  area  and  having  a  dip  of  10  ft.     Assum- 
ing the  load  on  the  platform  to  be  112  Ibs.  per  square  foot,  find   the 
greatest  pull  on  the  cables,  their   sectional    area,  length,  and    weight. 
(Safe  stress  =  8960  Ibs.  per  square  inch  ;  specific  weight  of  cable  =  480 
Ibs.  per  cubic  foot.) 

Ans.  H  =  —-=T=  56,000 Ibs.;  area  =6.73  sq.  in.; 

4/29 

length  =  io2f  ft.;  weight  =  2302.65  Ibs. 

10.  Find  the  depression  in  the  cables  in  the  last  question  due  to  an 
increment  of  length  under  a  change  of  60°  F.  from  the  mean  temperature. 
(Coefficient  of  expansion  =i  -f-  144000.)  Ans.  .0802  ft. 

n.  Each  side  of  the  platform  of  a  suspension-bridge  for  a  span  of  100 
ft.  is  carried  by  nine  equidistant  suspenders.  Design  a  stiffening  truss  for 
a  live  load  of  1000  Ibs.  per  lineal  foot,  and  determine  the  pull  upon  the 
suspenders  due  to  the  live  load  when  the  load  produces  (i)  an  absolute 
maximum  shear ;  (2)  an  absolute  maximum  bending  moment. 

Ans.  Max.  shear  =  6250  Ibs.;  max.  B.M.  =  92,592^  ft.-lbs.;  pull 
on  suspender  =  (i)  2777!  Ibs.,  (2)  =  185 iff  Ibs.  or  3703-1 rf  Ibs. 

12.  In  a  suspension-bridge  (recently  blown  down)  each  cable  was  de- 
signed to  carry  a  total  load  of  84  tons  (including  its  own  weight).  The 
distance  between  the  piers  =  1270  ft.;  the  deflection  of  the  cable  =  91  ft. 
Find  (a)  the  length  of  the  cable ;  (b)  the  pull  on  the  cable  at  the  piers 
and  at  the  lowest  point ;  (c)  the  amounts  by  which  these  pulls  are  changed 
by  a  variation  of  40°  F.  from  the  mean  temperature  ;  (</)  the  tension  in 
the  back-stays,  assuming  them  to  be  approximately  straight  and  inclined 
to  the  vertical  at  the  angle  whose  tangent  is  f. 


THEORY  OF  STRUCTURES. 

T 
Ans.— (a)  1287.4  ft.;  (£)  H— =  146^  tons ;  (<r)  depression  due 

to  change  of  temp.  =  .936  ft.  and  amount  of  change  in  H  =  ' H 

=  ii  tons,  in   T  =1.45  tons;    (d)  394.55   tons,  neglecting   pier 
friction. 

13.  The  platform  of  the  bridge  in  the  preceding  question  was  hung 
from  the  cables  by  means  of  480  suspenders  (240  on   each  side).     Find 
the  pull  on  each  suspender  and  the  total  length  of  the  suspenders,  the 
lowest  point  of  a  cable  being  14  ft.  above  the  platform. 

Ans.  .35  ton  ;   10,565^!-!  ft. 

14.  A  suspension-bridge  has  a  dip  of  10  ft.  and  a  span  of  300  ft.    Find 
the  increase  of  dip  due  to  a  change  of  100°  F.  from  the  mean  tempera- 
ture, the  coefficient  of  expansion  being  .00125  per  180°  F. 

Ans.  1.17  ft. 

Also,  find   the   corresponding   flange   stress  in  the  stiffening  truss, 
which  is  12^  ft.  deep,  the  coefficient  of  elasticity  being  8000  tons. 

Ans.  6.24  tons. 

15.  The  ends  of  a  cable  are  attached  to  saddles  free  to  move  horizon- 
tally.     If  Aa  is  the  horizontal  movement  of  each  saddle  due  to  the  ex- 
pansion of  the  cables  in  the  side  spans,  and  if  AS  is  the  extension  of  the 
chain  between  the  two  saddles,  show  that  the  increment  of  the  dip  (fi)  is 
approximately 

$h~a\ 

16.  The  platform  of  a  suspension-bridge  of  150  ft.  span  is  suspended 
from  the  two  cables  by  88  vertical  rods  (44  on  each  side)  ;  the  dip  of  the 
cables  is  15  ft.;  there  are  two  stiffening  trusses  ;  the  dead  weight  is  2240 
Ibs.  per  lineal  foot,  of  which  one-half  is  divided  equally  between  the  two 
piers.     Find  the  stresses  at  the  middle  and  ends  of  the  cables  when  a 
uniformly  distributed  load  of  78,750  Ibs.  covers  one  half  of  the  bridge. 
Also,  find  the  maximum  shears  and  bending  moments  to  which  the  stiff- 
ening trusses  are  subjected  when  a  live  load  of  1050  Ibs.  per  lineal  foot 
crosses  the  bridge. 

Ans.  Pull  on  suspender  =  2741$  Ibs.;  H=   -  -  T=  203,437^^5. 

^29 

Max.  shear  on  each  truss  at  centre  and  due  to  78,750  Ibs. 
=  9843^  Ibs.  =  that  due  to  10*50  Ibs.  per  lineal  foot. 

Max.  B.  M.  due  to  78,750  Ibs.  is  at  centre  of  loaded  and  un- 
loaded halves  and  =  184,570^  ft. -Ibs. 

Abs.  max.  B.M.  due  to  1050  Ibs.  per  lineal  foot  is  at  points 
of  trisection  and  ==  218,750  ft. -Ibs. 


EXAMPLES.  737 

17.  Solve  the  preceding  question  when  the  trusses  are  hinged  at  the 
centre. 

Ans.  Pull  on  suspender  =  2741!  Ibs.  ;   H  —  --=  T  =  1  54,2  1  Sf  Ibs. 


Max.  shear  due  to  78,750  Ibs.  =  9843!  Ibs.  at  centre  of 
span  and  at  end  of  loaded  half  of  bridge  ;  max.  shears 
due  to  1050  Ibs.  per  lineal  £001  =  13,125,  5906^,  4921^, 
&3°5TZ8>  anc*  9843!  Ibs.  at  ends  of  the  half  truss  and  at 
the  points  dividing  the  half  span  into  four  equal  seg- 
ments. 

Max.  B.  M.  due  to  78,750  Ibs.  is  at  centre  of  half  truss  and 
=  184,570^-  ft.  -Ibs.  Max.  B.  M.  due  to  1050  Ibs.  per  lineal 
foot  =176,  1  80  ff|,  221,484!,  and  153,808-^1  ft.-lbs.  at  points 
dividing  the  half  truss  into  four  equal  segments. 

1  8.  Show  that  the  total  extension  of  a  cable  of  uniform  sectional  area 
A  under  a  uniformly  distributed  load  of  intensity  iu  is 


wl 


3    /         16  </3\ 

-    I    +    — -TTl, 


8£.M  3 

/  being  the  span  and  d  the  dip. 

19.  The  dead  weight  of  a  suspension-bridge  of  1600  ft.  span  is  \  ton 
per  lineal  foot;  the  dip  = .     Find  the  greatest  and  least  pulls  upon 

one  of  the  chains.  The  ends  of  the  chains  are  attached  to  saddles  on 
rollers  on  the  top  of  piers  50  ft.  high,  and  the  back-stays  are  anchored 
50  ft.  from  the  foot  of  each  pier.  Find  the  load  upon  the  piers  and  the 
pull  upon  the  anchorage. 

Ans.  255  tons  ;  243!  tons  ;  637^  tons  ;  344.6  tons. 

20.  A  bridge  444  ft.  long  consists  of  a  central  span  of  180  ft.  and  two 
side  spans  each  of  132  ft.  ;  each  side  of  the  platform  is  suspended  by 
vertical  rods  from  two  iron-wire  cables ;  each  pair  of  cables  passes  over 
two  masonry  abutments  and  two  piers,  the  former  being  24  ft.  and  the 

atter  39  ft.  above  the  surface  of  the  ground  ;  the  lowest  point  of  the 
cables  in  each  span  is  19  ft.  above  the  ground  surface  ;  at  the  abutments 
the  cables  are  connected  with  straight  wrought-iron  chains,  by  means  of 
which  they  are  attached  to  anchorages  at  a  horizontal  distance  of  66  ft. 
from  the  foot  of  each  abutment ;  the  dead  weight  of  the  bridge  is  3500 
Ibs.  per  lineal  foot,  and  the  bridge  is  covered  with  a  proof  load  of  4500 
Ibs.  per  lineal  foot.  Determine — 

(a)  The  stresses  in  the  cables  at  the  points  of  support  and  at  the 
lowest  points. 

(d}  The  dimensions  and  weights  of  the  cables  (i)  if  of  uniform  sec- 


73 8  THEORY  OF  STRUCTURES. 

tion  throughout;  (2)  if  each  section  is  proportioned  to  the  pull  across  it. 
(Unit  stress  =  14.958  Ibs.  per  square  inch.) 

(c)  The  alteration  in  the  length  of  the  cables  and  the  corresponding 
depression  of  the  platform  at  the  centre  of  each  span,  due  (i)  to  a  change 
of  60°  F.  from  the  mean  temperature  ;    (2)  to  the  total  load,  E  being 
30,000,000  Ibs. 

(d)  The  pressure  and  bending  moment  at  the  foot  of  pier. 

(e)  The  mass  of   masonry  in  the  anchorage  necessary  to  resist  the 
tendency  to  overturning  and  to  horizontal  displacement. 

Data. — Weight  of  masonry  per  cubic  foot  =  128  Ibs.  ;  safe  compres- 
sive  stress  per  square  foot  =  12,000  Ibs.  ;  coefficient  of  friction  =76; 
deviation  of  centre  of  pressure  in  base  of  pier  from  centre  of  figure  =  f 
x  thickness  of  base. 

Am.— (a)  Side  span,  Ti—z=  —  Hl  —  387,200  Ibs.  =  T3-^= ,     T, 
1/509  4/146 

and  7a  being  the  tensions  in  a  cable  at  summits  of 

Q 

low  and  high  piers,  respectively;  centre  span,  T-=-. - 

V97 

=  405,000  Ibs.  =  H,  T  being  tension  at  summit  of  high 
pier. 

(£)  Side  span  :  Length  =  135/3  ft-  ;  sect,  area  at  summit 
of  high  pier  =  28.43  scl-  m-  5  weight  if  of  uniform  section 
=  12,834  Ibs.,  if  proportioned  to  pull  —  11,710  Ibs. 
Centre  span  :  Length  =  i85ff  ft. ;  sect,  area  at  summit 
of  pier  =  29.6  sq.  in.;  weight  if  of  uniform  section 
=  18,361  Ibs.,  if  proportioned  to  pull  =  17,267  Ibs. 
(c)  (i)  .0594  ft.  for  side  span  and  .0775  ft-  f°r  centre  span  ; 

(2)  .0675  "     "       "        "       "     -0927  "     " 
(d)  High  pier:    Overturning   moment  =  694,200  ft. -Ibs. ; 
bearing    area    at    summit  =  uSf    sq.  ft.;   thickness 
=  8  ft.  ;    uniform  width  =  14!  ft.;   thickness  of  base 
=  10.7  ft. ;  weight  of  pier  =  692,348.8  Ibs.  ;  total  pres- 
sure on  base  =  2,116,348.8  Ibs. 
(<?)  Weight  to  resist   upward  pull  =  29,333^  Ibs. ;  weight 

to  resist  horizontal  displacement  =  509,474  Ibs. 

21.  In  the  preceding  question,  if  the  piers  are  wrought-iron  oscillat- 
ing columns,  and  if  equilibrium,  under  an  unequally  distributed  load,  is 
maintained  by  connecting  the  heads  of  the  columns  with  each  other  and 
with  the  abutments  by  iron-wire  stays,  determine  the  proper  dimensions 
of  the  stays,  assuming  them  approximately  straight.  Assume  that  the 
proof  load  covers  (a)  a  side  span  ;  (b)  two  side  spans ;  (c)  the  centre  span. 
Ans. — (a)  Pull  on  stays  in  centre  span  =  840,050  Ibs. 

(b)      "    J'       "       "       "  "     =  double  that  in  (a). 

00      "      "       "       "   side  span      =  948,432  Ibs. 


EXAMPLES.  739 

22.  A  floating  landing-stage  is  held  in  position  by  a  number  of  4^- 
iti.  steel-wire  cables  anchored  to  the  shore,  a  shoreward  movement  being 
prevented  by  rigid  iron  booms,  pivoted  at  the  ends  and  stretching  from 
shore  to  stage.  The  difference  of  level  between  the  shore  and  stage  at- 
tachments of  the  cables  is  50  ft.,  and  the  horizontal  distance  between 
these  points  is  150  ft.  The  horizontal  pull  upon  each  cable  is  1360  Ibs. 
Find  the  length  of  the  cable,  and  the  tensions  at  the  points  of  attach- 
ment. (Weight  of  cable  =  490  Ibs.  per  cubic  foot ;  form  of  cable  a  com- 
mon catenary.)  Am.  342.82ft.;  12,267.2  Ibs.  and  10,132  Ibs. 


CHAPTER    XIII. 
ARCHES   AND   ARCHED    RIBS. 

I.  AN   arch    may  be   constructed  of   masonry,  brickwork, 
timber,  or  metal. 


FIG.  478. 

In  the  figure  ABCD  represents  the  profile  of  an  arch. '  The 
under  surface  AD  is  called  the  soffit  or  intrados.  The  upper 
surface  BC  is  sometimes  improperly  called  the  extrados.  The 
highest  point  A"  of  the  soffit  is  the  crown  or  key  of  the  arch. 
The  springings  or  skewbacks  are  the  surfaces  y^.5,  DC  from 
which  the  arch  springs,  and  the  haunches  are  the  portions  of 
the  arch  half-way  between  the  springings  and  the  crown. 
Upon  each  of  the  arch  faces  stands  a  spandril  wall,  and  the 
space  between  these  two  external  spandrils  may  be  occupied 
by  a  series  of  internal  spandrils  spaced  at  definite  distances 
apart,  or  may  be  filled  up  to  a  certain  level  with  masonry  (i.e., 
backing]  and  above  that  with  ordinary  ballast  or  other  rough 
material  (i.e.,  filling). 

A  masonry  arch  consists  of  courses  of  wedge-shaped  blocks 
with  the  bed-joints  perpendicular,  or  nearly  so,  to  the  soffit. 

740 


EQUILIBRATED   POLYGON  AND   LINE   OF  RESISTANCE.     74! 

The  blocks  are  called  voussoirs,  and  the  voussoirs  at  the  crown 
are  the  keystones  of  the  arch. 

A  brick  arch  is  usually  built  in  a  number  of  rings. 

Consider  the  portion  of  the  arch  bounded  by  the  vertical 
plane  KE  at  the  key  and  by  the  plane  AB. 

It  is  kept  in  equilibrium  by  the  reaction  R  at  KE,  the  reac- 
tion Rl  at  AB,  and  the  weight  F,  of  the  portion  under  con- 
sideration and  its  superincumbent  load. 

Let  S  and  T  be  the  points  of  application  of  R^  and  R, 
respectively. 

Let  the  directions  of  R,  and  R  intersect  in  a  point.  The 
direction  of  F,  must  also  pass  through  the  same  point. 

Taking  moments  about  5, 


pl  and  jj/j  being  the  perpendicular  distances  of  the  directions  of 
R  and  F,  from  S,  respectively. 

Similarly,  the  portion  KECD  of  the  arch  gives  the  equation 


F9  being  the  weight  to  which  it  is  subjected,  and  /a,  y^  the 
perpendicular  distances  of  the  directions  of  R  and  F2  from  the 
point  of  application  Fof  the  reaction  at  the  plane  DC. 

If  the  arch  and  the  loading  are  symmetrical  with  respect 
to  the  plane  KE, 

Yl  =  Y9t    yi=y^y     and  therefore  /,  —  /2. 

Hence  the  direction  of  R  will  be  horizontal,  which  might  have 
been  inferred  by  reason  of  the  symmetry. 

The  magnitudes  of  the  reactions  are  indeterminate,  as  the 
positions  of  the  points  of  application  (5,  T,  F)  are  arbitrary, 
and  can  only  be  fixed  by  a  knowledge  of  the  law  of  the  varia- 
tion of  the  stress  in  the  material  at  the  bounding  planes  AB, 
KE. 

2.  Equilibrated  Polygon  and  Line  of  Resistance.— 
Suppose  an  arch  divided  into  a  number  of  elementary  portions 


THEORY  OF   STRUCTURES. 


ke' ,  k' e"  .  .  .  (e.g.,  the  voussoirs  of  a  masonry  arch)  by  a  series 
of  joints  ke,  k'e'  .  .  . 


FIG.  479. 


FIG.  480. 


Let  Wiy  W^j  „  .  .  be  the  loads  directly  supported  by  the 
several  portions.  These  loads  generally  consist  of  the  weight 
of  a  portion  (e.g.,  ke')  +  the  weight  of  the  superincumbent 
mass  -\-  the  load  upon  the  overlying  roadway  ;  the  lines  of 
action  of  the  loads  are,  therefore,  nearly  always  vertical. 

Each  elementary  portion  may  be  considered  as  acted  upon 
and  kept  in  equilibrium  by  three  forces,  viz.,  the  external  load 
and  the  pressures  at  the  joints.  If  the  pressure  and  its  point 
of  application  at  any  given  joint  have  been  determined,  the 
pressures  and  the  corresponding  points  of  application  at  the 
other  joints  may  also  be  found. 

For,  let  i  2  34  ...  be  the  line  of  loads,  so  that  12=  W^ 

23=  W, 

Assume  that  the  pressure  P  and  its  point  of  application  r 
at  any  given  joint  ke  are  known. 

Draw  o  I  to  represent  P  in  direction  and  magnitude. 

Then  02  evidently  represents  the  resultant  of  P  and  Wl  in 
direction  and  magnitude,  and  this  resultant  must  be  equal  and 
opposite  to  the  pressure  Pl  at  the  joint  k'e' . 

Hence,  a  line  n'n  drawn  through  n,  the  intersection  of  P 
and  Wlt  parallel  to  20,  is  the  direction  of  the  pressure  Plt  and 
intersects  k'e'  in  the  point  of  application  r'  of  P, 

Again,  o  3  represents  the  resultant  of  /\  and  W^  in  direc- 
tion and  magnitude,  and  this  resultant  must  be  equal  and 
opposite  to  the  pressure  P9  at  the  joint  k"e" . 

The  line  n"ri  drawn   through  «',  the  intersection  of  P^  and 


EQUILIBRATED   POLYGON  AND  LINE   OF  RESISTANCE.     743 


W2,  parallel  to  30,  is  the  direction  of  the  pressure  P9  and  inter- 
sects k"e"  in. the  point  of  application  r"  of  Pt. 

Proceeding  in  this  manner,  a  series  of  points  of  application 
or  centres  of  resistance  r',  r",  r'" ,  .  .  .  may  be  found,  the 
corresponding  pressures  being  represented  by  02,  03^04,  .  .  . 

The  polygon  of  pressures  formed  by  the  lines  of  action  of 
P,  Plt  P^,  .  .  .  is  termed  an  equilibrated  polygon,  and  is  a  funic- 
ular polygon  of  the  loads  upon  the  several  portions. 

The  polygon  formed  by  joining  the  points  r,  rf,  r" ,  .  .  . 
successively,  is  called  the  line  of  resistance. 

In  the  limit,  when  the  joints  are  supposed  indefinitely  near, 
these  polygons  become  curves,  the  curve  in  the  case  of  the 
equilibrated  polygon  being  known  as  a  linear  arch. 

The  two  curves  may,  without  sensible  error,  be  supposed 
identical,  and  they  will  exactly  coincide  if  the  joints  (of  course 
imaginary  in  such  a  case)  are  made  parallel  to  the  lines  of 
action  of  the  external  loads.  This  may  be  easily  proved  as 
follows : 

Let  the  figure  represent  a  portion  of  an  arch  bounded  by 
the  joints  (imaginary)  KE,  MN  parallel  to  the  lines  of  action 
of  the  external  loads,  which  will  be  assumed  vertical. 

Reduce  the  superincumbent  loads  to  an  equivalent  mass  of 
arch  material. 

Let  h,  e.g.,  be  the  depth  of  material  of  specific  weight  wl , 
overlying  the  arch  at  any  given 
point,  and  let  Q  be  the  load  per 
unit  of  area  of  roadway. 

Also,   let   iv  be   the  specific 
weight  of  the  arch  material. 

Then  x,  the  equivalent  depth, 
is  given  by 


wx  =  wjt  -\-  Q. 


FIG.  481 


If  the  value  of  x  is  deter- 
mined at  different  points  along  the  arch,  a  profiU,  en  may  be 
obtained  defining  a  mass  ENne  of  arch  material  which  may  be 
substituted  for  the  superincumbent  load.  Denote  the  weight 
of  the  mass  MKen  by  W. 


744  THEORY  OF  STRUCTURES. 

Let  the  pressure  P  and  its  point  of  application  O  at  the 
joint  KE  be  given. 

Take  O  as  the  origin,  the  line  OA  in  the  direction  of  P  as 
the  axis  of  x,  and  the  vertical  through  O  as  the  axis  of  jy,  and 
let  0  be  the  angle  between  the  two  axes. 

Let  the  lines  of  action  of  P  and  W  intersect  in  G.  The 
line  of  action  of  their  resultant  will  intersect  MN  m  the  centre 
of  resistance  Or 

Let  X,  Y  be  the  co-ordinates  of  O,. 

Let  z  be  the  depth  of  an  elementary  slice  of  thickness  dx, 
parallel  to  OK  at  any  abscissa  x.  Its  weight  —  wzdx  sin  0. 
Then 


W.  OG  =       wzdx  smB.x=  W(X  -  AG\ 

But  -vp=  ~TQ  —  ~~p~>  since  tne  triangle  AGO  is  evidently  a 

triangle  of  forces  for  the  forces  acting  upon  the  mass  under 
consideration. 
Also, 

W  =    /  wzdx  .  sin  0. 
.'.    Cwzxdx  .  sin  0  =  WX  -  W~Y=  X  ffvz  sin  Odx  -  PY. 

t/o  W  t/o 

This  is  the  equation  to  the  line  of  resistance. 
Taking  the  differential  of  this  equation, 

wz'X  sin  BdX  =  Xwz'  sin  BdX  +  WdX  -  PdY9 
z'  being  the  depth  corresponding  to  the  abscissa  X. 
dY       W       AO 


Thus  the  tangents  to  the  curve  of  pressures  and  to  the 
curve  of  centres  of  pressure  at  any  given  point  coincide,  and 
the  curves  must  therefore  also  coincide. 


CONDITIONS   OF  EQUILIBRIUM.  745 

3.  Conditions  of  Equilibrium. — Let  the  figure  represent 
a  portion  of  an  arch  of  thickness  unity,  between  any  two  bed- 
joints  (real  or  imaginary]  MN,  PQ. 

Let  W  be  its  weight  together  with  that  of  the  superincum- 
bent load.     Let  the  direction   of 
the  reaction  R'  at  the  joint  MN 
intersect  MN  in  m  and  the  direc- 
tion of  J^in  n.     For  equilibrium, 
the  reaction  R"  at  the  joint  PQQ. 
must  also  pass  through  n.    Let  its 
direction  intersect  PQ  in  O.     In          O  +W 

order  that  the  equilibrium  may  be 
stable,  three   conditions  must   be  HIG'  48z< 

fulfilled,  viz. : 

First.  The  point  O  must  lie  between  P  and  Q,  so  that  there 
may  be  no  tendency  to  turn  about  the  edges  Pand  Q. 

Second.  There  must  be  no  sliding  along  PQ,  and  therefore 
the  angle  between  the  direction  of  R"  and  the  normal  to  PQ 
must  not  exceed  the  angle  of  friction  of  the  material  of  which 
the  arch  is  composed. 

N.B. — The  angle  of  friction  for  stone  upon  stone  is  about 

30°. 

Third.  The  maximum  intensity  of  stress  at  any  point  in  PQ 
must  not  exceed  the  safe  resistance  of  the  material. 

Further,  the  stress  should  not  change  in  character,  in  the 
case  of  masonry  and  brick  arches,  but  should  be  a  compression 
at  every  point,  as  these  materials  are  not  suited  to  withstand 
tensile  forces. 

The  best  position  for  O  would  be  the  middle  point  of  PQ, 
as  the  pressure  would  then  be  uniformly  distributed  over  the 
area  PQ.  It  is,  however,  impracticable  to  insure  such  a  dis- 
tribution, and  it  has  been  sometimes  assumed  that  the  stress 
varies  uniformly, 

With  this  assumption,  let  ^Vbe  the  normal  component  of  R". 

Let /be  the  maximum  compressive  stress,  i.e.,  the  stress  at 
the  most  compressed  edge,  e.g.,  P. 

Let  OS  —  q  .  PQ,  S  being  the  middle  point  of  PQ,  and  q  a 
coefficient  whose  value  is  to  be  determined. 


746  THEORY   OF  STRUCTURES. 

PO 
Then  if  PO  <  —  , 


N-f'PQ 


and  in  the  limit  when  PO  =  —  ,  i.e.,  when  the  intensity  of 
stress  varies  uniformly  from  /at  Pto  nil  at  Q, 

?=*     and     &** 

(See  Art.  16,  Chap.  IV.) 

Similarly,  if  Q  is  the  most  compressed  edge,  the  limiting 
position  of  O,  the  centre  of  resistance  or  pressure,  is  at  a  point 

PQ 
O'  denned  by  QO'  -  ~. 

Hence,  as  there  should  be  no  tendency  on  the  part  of  the 
joints  to  open  at  either  edge,  it  is  inferred  that  PO  or  QO' 

PQ 
should  be  >  —  —  ,  i.e.,  that  the  point   O  should  lie  within  the 

0 

middle  third  of  the  joint. 

Experience,  however,  shows  that  the  "  middle-third  " 
theory  cannot  be  accepted  as  a  solution  of  the  problem  of 
arch  stability,  and  that  its  chief  use  is  to  indicate  the  proper 
dimensions  of  the  abutments.  Joint  cracks  are  to  be  found  in 
more  than  90$  of  the  arches  actually  constructed,  and  cases 
may  be  instanced  in  which  the  joints  have  opened  so  widely 
that  the  whole  of  the  thrust  is  transmitted  through  the  edges. 
In  Telford's  masonry  arch  over  the  Severn,  of  150  ft.  span, 
Baker  discovered  that  there  had  been  a  settlement  (15  in.) 
sufficient  to  induce  a  slight  reverse  curvature  at  the  crown  of 
the  soffit.  Again,  the  position  of  the  centre  of  pressure  at  a 
joint  is  indeterminate,  and  it  is  therefore  impossible  as  well  as 
useless  to  make  any  calculations  as  to  the  maximum  intensity 
of  stress  due  to  the  pressure  at  the  joint.  What  seems  to 


JOINT  OF  RUPTURE.  747 

happen  in  practice  is,  that  the  straining  at  the  joints  generally 
exceeds  the  limit  of  elasticity,  and  that  the  pressure  is  uni- 
formly distributed  for  a  certain  distance  on  each  side  of  the 
curve  of  pressures.  Thus,  the  proper  dimensions  of  a  stable 
arch  are  usually  determined  by  empirical  rules  which  have 
been  deduced  as  the  results  of  experience.  For  example, 
Baker  makes  the  following  statement : 

Let  T  be  the  thrust  in  tons  or  pounds  per  lineal  foot  of 
width  of  arch. 

Let  f  be  the  safe  working  stress  in  tons  or  pounds  per 
square  foot. 

An  arch  will  be  stable  if  an  ideal  arch,  with  its  bounding 

i  T 

surfaces  at  a  minimum  distance  of  —  —  from  the  curve  of  pres- 
sures, can  be  traced  so  as  to  lie  within  the  actual  arch.  An 
advance  would  be  made  towards  a  more  correct  theory  if  it 
were  possible  to  introduce  into  the  question,  the  elasticity  and 
compressibility  of  the  materials  of  construction.  These  ele- 
ments, however,  vary  between  such  wide  limits  that  no 
reliance  can  be  placed  upon  the  stresses  derivable  from  their 
values. 

4.  Joint  of  Rupture. — Let  I  2,  3  4  be  the  bounding  surfaces 
between  which  the  curve  of  pressures  must  lie,  and  let  4  be 


2 

FIG.  483. 

the  centre  of  pressure  at  the  crown.  A  series  of  curves  of 
pressure  may  be  drawn  for  the  same  given  load,  but  with 
different  values  of  the  horizontal  thrust  h. 

Let  AfXy  be  that  particular  curve  which  for  a  value  //of  the 
horizontal  thrust  is  tangent  to  the  surface  I  2  at  x ;  the  joint  at 
x  is  called  the  joint  of  rupture. 

The    angle    which    the   joint  of    rupture    makes  with    the 


748 


THEORY  OF  STRUCTURES. 


horizontal  is  about  30°  in  semicircular  and  45°  in  elliptic 
arches. 

The  position  of  the  joint  in  any  given  arch  may  be  tenta- 
tively found  as  follows : 

Lety  be  any  joint  in  the  surface  I  2. 

Let  Wbe  the  weight  upon  the  arch  between  /"and  I. 

Let  X  be  the  horizontal  distance  between  J  and  the  centre 
of  gravity  of  W. 

Let  Y  be  the  vertical  distance  between  J  and  4. 

It  will  also  be  assumed  that  the  thrust  at  4  is  horizontal. 

If  the  curve  of  pressure  be  now  supposed  to  pass  through 
J,  the  corresponding  value  of  the  horizontal  thrust  h  is  given 
by 

kY=  WX. 

By  means  of  this  equation,  values  of  h  may  be  calculated 
for  a  number  of  joints  in  the  neighborhood  of  the  haunch,  and 
the  greatest  of  these  values  will  be  the  horizontal  thrust  H  for 
the  joint  x.  This  is  evident,  as  the  curve  of  pressure  for  a 
smaller  value  of  h  must  necessarily  fall  below  ^xy. 

When  this  happens,  the  joints  will  tend  to  open  at  the 
lower  edge  of  the  joint  I  4  and  at  the  upper  edges  of  the  joints 
at  x  and  at  2  3,  so  that  the  arch  may  sink  at  the  crown  and 
spread,  unless  the  abutments  and  the  lower  portions  of  the 
arch  are  massive  enough  to  counteract  this  tendency. 

If  the  curve  of  pressure  fall  above  A^xy,  an  amount  of  back- 
ing sufficient  to  transmit  the  thrust  to  the  abutments  must  be 
provided.  The  same  result  may  be  attained  by  a  uniform  in- 
crease in  the  thickness  of  the  arch  ring,  or  by  a  gradual  increase 
from  the  crown  to  the  abutments. 

For  example,  the  upper  sur- 
face (extrados)  of  the  ring  for  an 
arch  with  a  semicircular  soffit 
A  KB,  having  its  centre  at  O,  may 
be  delineated  in  the  following 
manner: 

^ Let  x  define  the  joint  of  rup- 

ture in  the  soffit ;     then  AOx  —  30°. 


\ 


MINIMUM    THICKNESS   OF  ABUTMENT. 


749 


be 

Y 


the 


In  Ox  produced  take  xxr  =  2  X  KD,  KD  being  the  thick- 
ness at  the  crown. 

The  arc  Dx'  of  a  circle  struck  from  a  centre  in  DO  pro- 
duced may  be  taken  as  a  part  of  the  upper  boundary  of  the 
ring,  and  the  remainder  may  be  completed  by  the  tangent  at 
xl  to  the  arc  Dx' . 

5.  Minimum   Thickness  of  Abutment.— Let 
resultant  thrust  at  the  horizontal  joint  BC  of  a 
rectangular  abutment  ABCD. 

Let  y  be  the  distance  of  its  point  of  applica-  C 
tion  from  B. 

Let  //  and  V  be  the  horizontal  and  vertical 
components  of  T. 

Let  w  be  the  specific  weight  of  the  material 
in  the  abutment. 

Let  h  be  the  height  AB  of  the  abutment. 

Let  t  be  the  width  AD  of  the  abutment.  FIG.  485. 

In  order  that  there  may  be  no  tendency  to  turn  about  the 
toe  D,  the  moment  of  the  weight  of  the  abutment  with  respect 
to  D  plus  the  moment  of  V  with  respect  to  D  must  be  greater 
than  the  moment  of  H  with  respect  to  D.  Or, 


or 


-  +  V(t-y)>  Hh, 


H      \  I2— 4-  2V     4-    V* 
h  >  V    W~*  Wk-     ufk* 


This  relation   must  hold  good  whatever  the  height  of  the 
abutment  may  be  ;  and  if  h  is  made  equal  to  oo , 


which  defines  a  minimum  limit  for  the  thickness  of  the  abut- 
ment. 


750  THEORY  OF  STRUCTURES. 

6.  Empirical  Formulae. — In  practice    the  thickness  /at 
the  crown  is  often  found  in  terms  of  s,  the  span,  or  in  terms  of 
p,  the  radius  of  curvature  at  the  crown,  from  the  formulae 

t  =  c  Vs,     or     t  =  Vcp, 
t,  s,  and  p  being  all  in  feet,  and  c  being  a  constant. 

According  to  Dupuit,  /  =  .36  Vs  for  a  full  arch ; 

/  =  .27  Vs  for  a  segmental  arch. 
According  to  Rankine,    =  l/.i  2p  for  a  single  arch  ; 

/  =  V.ijp  for  an  arch  of  a  series. 

7.  Examples  of  Linear  Arches,  or  Curves  of  Pressure. 
\(a)  Linear  Arch  in  the  Form  of  a  Parabola. — Suppose  that 

the  cable  in  Art.  4,  Chap.  XII,  Case  B,  is  exactly  inverted, 
and  that  it  is  stiffened  in  such  a  manner  as  to  resist  distortion. 
Suppose  also  that  the  load  still  remains  a  uniformly  distributed 
weight  of  intensity  w  per  horizontal  unit  of  length.  A  thrust 
will  now  be  developed  at  every  point  of  the  inverted  cable 
equal  to  the  tension  at  the  corresponding  point  of  the  original 
cable.  Thus  the  inverted  parabola  is  a  linear  arch  suitable  for 
a  real  arch  which  has  to  support  a  load  of  intensity  w  per 
horizontal  unit  of  length. 

The  horizontal  thrust  at  the  crown  =  H  =  wp, 

p  being  the  radius  of  curvature  at  the  crown. 

(fr)  Linear  Arch  in  the  Form  of  a  Catenary.      Transformed 
Catenary. — If  the  cable  in  Art.  4,  Chap.  XIT,  Case  A,  is  in- 
N         T      o  N      verted  and  stiffened  as  before,  a  linear 

arch   is   obtained   suitable   for"  a    real 
arch  which  has  to  support  a  load  dis- 
tributed  in  such   a   manner  that  the 
weight  upon  any  portion  AP  is  pro- 
p,   portional  to  the  length  of  AP,  and  is 
in  fact  =ps.     The  area  OAPN '  —  ms. 
Thus,  a  lamina  of  thickness  unity 

and  specific  weight  w,  bounded  by  the  curve  AP,  the  directrix 
ON,  and  the  verticals  AO,  PN,  weighs  wms,  and  may  be  taken 


EXAMPLES  OF  LINEAR  ARCHES.  75  r 

to  represent  the  load  upon  the  arch  if  wms  —  ps,  i.e.,  if  wm  =  J>, 
i.e.,  if  the  weight  of  m  units  of  the  lamina  is  w. 

The  horizontal  thrust  at  the  crown  —  H  —  wm  =  wp, 

the  radius  of  curvature  (p)  at  the  crown  being  equal  to  m. 

A  disadvantage  attached  to  a  linear  arch  in  the  form  of  a 
catenary  lies  in  the  fact  that  only  one  catenary  can  pass 
through  two  given  points,  while,  in  practice,  it  is  often  neces- 
sary that  an  arch  shall  pass  through  three  ^points  in  order  to 
meet  the  requirements  of  a  given  rise  and  span.  This  difficulty 
may  be  obviated  by  the  use  of  the  transformed  catenary. 

Upon  the  lamina  PAPNN  as  base,  erect  a  solid,  with  its 
horizontal  sections  all  the  same,  and,  for  simplicity,  with  its 
generating  line  perpendicular  to  the  base. 

Cut  this  solid  by  a  plane  through  NN  inclined  at  any  re- 
quired angle  to  the  base.  The  intersection  of  the  plane  and 
solid  will  define  a  transformed  catenary  P'A'P',  or  a  new  linear 
arch,  and  the  shape  of  a  new  lamina  P'A'P'NN,  under  which 
the  arch  will  be  balanced.  This  is  evident,  as  the  new  arch  and 
lamina  are  merely  parallel  projections  of  the  original. 

The  projections  of  horizontal  lines  will  remain  the  same  in 
length. 

The  projections  of  vertical  lines  will  be  c  times  the  lengths 
of  the  lines  from  which  they  are  projected,  c  being  the  secant 
of  the  angle  made  by  the  cutting  plane  with  the  base. 

Let  .*•,  Y  be  the  co-ordinates  of  any  point  P'  of  the  trans- 
formed catenary. 

Let  x,  y  be  the  co-ordinates  of  the  corresponding  point  P 
in  the  catenary  proper. 


Then 

YPW  A'Q      M 


_       _ 
=        ~ 


y  ~    PN  ~  AO 

The  equation  to  the  catenary  proper  is 


(2) 


75  ^  THEORY  OF  STRUCTURES. 

Substituting  in  the  last  equation  the  value  of  y  given  by  eq.  (i), 


which  is  the  equation  to  the  transformed  catenary. 

With  this  form  of  Linear  arch  the  depths  M  over  the  crown 
and  Y  over  the  springings,  for  a  span  2x,  may  be  assumed,  and 
the  corresponding  value  of  m  determined  from  eq.  (3). 

It  is  convenient,  in  calculating  m,  to  write  eq.  (3)  in  the  form 


-'-     ...     (4) 

The  slope  i'  at  P'  is  given  by 

dY      Ml*-        -c\       Ms 
tarn    =  —  y—  =  —  I*    —  e       1  =  —  —, 
dx       2m\  I        m 

s  being  the  length  AP  of  the  catenary  proper,  corresponding 
to  the  length  A'P'  of  the  transformed  catenary. 


'P'N=     C 

<J  o 


The  area  OA'P'N=  Ydx  =         -e    -  e    m    =  Ms. 


The  triangle  P'  TN  is  a  triangle  of  forces  for  the  portion 
A'P'. 

The  triangle  PTN  is  a  triangle  of  forces  for  the  portion  AP. 

(The  tangents  at  P  and  P'  must  evidently  intersect  ON  in 
the  same  point  T.) 

Let  H'  be  the  horizontal  thrust  at  A',  //being  that  at  A. 

Let  P'  be  the  weight  upon  A'P't  P  being  that  upon  AP. 

Let  R'  be  the  thrust  at  P'. 

Then 

P'  __  area  OA'P'N  _  Ms  _M 
~P'     -  area  OAPN  ~~  ~ms~~^ny 


EXAMPLES   OF  LINEAR  ARCHES. 

and  hence 

pt  _  —p  —  —wms  —  wMs ; 
m          m 


H  =  P'  cot  i'  =  ivMs-jrf-  =  wm*  =  H\ 


/  J/V 

R'=  H'  sec  i'  =wm\/  i  -\ —  =  w 

V  m 


The  radius  of  curvature  p'  at  the  crown  =  -=-=  . 


. '.  H'  =  wMp'  —  H  —  wp, 

and  the  radius  of  the  "  catenary  proper  "  is  M  times  the  radius 
of  the  transformed  catenary. 

The  term  "  equilibrated  arch  "  has  generally  been  applied 
to  a  linear  arch  with  a  horizontal  extrados. 

(c)  Circular   and  Elliptic    Linear   Arches. — A   linear   arch 
which  has  to  support  an  external 
normal  pressure  of  uniform  inten- 
sity should  be  circular. 

Consider  an  indefinitely  small 
element  CD,  which  may  be  as- 
sumed to  be  approximately  j-~ 

"  KlG.    487. 

Let  the  direction  of  the  result- 
ant pressure  upon  CD,  viz.,/  .  CD,  make  an  angle  B  with  OB. 

Let  CE,  DE  be  the  vertical  and  horizontal  projections  of 
CD. 

The  angle  DCE  =  8. 

The  horizontal  component  of  p .  CD  =  p .  CD  cos  6  =  p .  CE. 

This  is  distributed  over  the  vertical  projection  CE. 

i       /.  the  .horizontal  intensity  of  pressure  =  /  .  CE  -j-  CE  =  p. 

Similarly,  it   may  be  shown   that   the  vertical    intensity  of 
pressure  — />. 


754 


THEORY  OF  STRUCTURES. 


Thus,  at  any  point  of  the  arch, 

the  horizontal  intensity  of  pressure 

=  vertical  intensity  =  normal  intensity  =/. 
Again,  the  total  horizontal  pressure  on  one-half  of  the  arch 
=  2(p.  CE)  =  p2(CE)  =  pr  =  H, 


and        the  total  vertical  pressure  on  one-half  of  the  arch 


=  2(p  .  DE)  = 


Hence,  at  any  point  of  the  arch  the  tangential  thrust  =  pr. 

Next,   upon  the  semicircle  as  base,  erect  a  semi-cylinder. 
Cut  the  latter  by  an  inclined  plane  drawn  through  a  line  in  the 


plane  of  the  base  parallel  to  OA.  The  intersection  of  the  cut- 
ting plane  and  the  semi-cylinder  is  the  semi-ellipse  B'AB',  in 
which  the  vertical  lines  are  unchanged  in  length,  while  the 
lengths  of  the  horizontal  lines  are  c  times  the  lengths  of  the 
corresponding  lines  in  the  semicircle,  c  being  the  secant  of  the 
angle  made  by  the  cutting  plane  with  the  base.  A  semi- 
elliptic  arch  is  thus  obtained,  and  the  forces  to  which  it  is  sub- 
jected are  parallel  projections  of  the  forces  acting  upon  the 
semicircular  arch. 

These  new  forces  are  in  equilibrium  (see  Corollary). 

Let  P'  =  the  total  vertical  pressure  upon  one-half  of  the 

arch  ; 

H'  —  the  total  horizontal  pressure  upon  one-half  of  the 
arch  ; 


EXAMPLES   OF  LINEAR  ARCHES,  755 

P 

py  =  vertical  intensity  of  pressure  =  ^-57  ; 

tft 

px'  =  horizontal  intensity  of  pressure  =  ~7=rr» 

Then 

Pf  =  P=H  =  pr;    ........     (i) 

P  P 


y  '-  OB'~  c.OB~~  cr  "" 
Hf  =  cH  =  cP  =  cP'-,  .......     (3) 


H' 
Hence,  by  eq.  (3), 


/_  ___ 

px  ~-  OA'  ~  OA  ~~  :  r    ~ 


*L         OB' 

~P~~        ~OA' 

or,  the  total  horizontal  and  vertical  thrusts  are  in  the  ratio  of 
the  axes  to  which  they  are  respectively  parallel,  and,  by  eqs. 
(2)  and  (4),  • 

A.' -I      OA" 

px'~S~  OB'*  ; 

or,  the  vertical  and  horizontal  intensities  of  pressure  are  in  the 
ratio  of  the  squares  of  the  axes  to  which  they  are  respectively 
parallel. 

Any  two  rectangular  axes  OG,  OK  in  the  circle  will  project 
into  a  pair  of  conjugate  radii  OG' ,  OK'  in  the  ellipse. 

Let  OG'  =  rlt  OK'  =  r2 ; 

Q  =  total  thrust  along  elliptic  arch  at  K\ 

E>  — —        a  «  «  «  «          «     /"• 

Then 

H      r  H      r 


THEORY  OF  STRUCTURES. 

or,  the  total  thrusts  along  an  elliptic  arch  at  the  extremities  of 
a  pair  of  conjugate  radii  are  in  the  ratio  of  the  radii  to  which 
they  are  respectively  parallel. 

The  preceding  results  show  that  an  elliptic  linear  arch  is 
suitable  for  a  load  distributed  in  such  a  manner  that  the  vertical 
and  horizontal  intensities  (eqs.  (2)  and  (4)  )  at  any  point  of  the 
arch  are  unequal,  but  are  uniform  in  direction  and  magnitude. 

Corollary.  —  It  can  be  easily  shown  that  the  projected  forces 
acting  upon  the  elliptic  arch  are  in  equilibrium. 

The  equations  of  equilibrium  for  the  forces  acting  upon  the 
circular  arch  may  be  written 


T  being  the  thrust  along  the  arch  at  the  point  xy,  and  X,  Y 
the  forces  acting  upon  the  arch  parallel  to  the  axes  of  x  and 
y,  respectively. 

If  T',  X',  V  be  the  corresponding  projected  forces, 

~  =  ~,     Xds  =  cX'ds',     Yds  =  Y'ds*. 


Hence,  the  above  equations  may  be  written 

d  j^p  cdx^  +  cX'ds'  =  o, 
and 

or 


and 

d\ 


(Tid£]  +  *'***• 

Hence,  the  forces  T1  ',  X',  and  Y'  are  also  in  equilibrium. 


EXAMPLES   OF  LINEAR  ARCHES. 

(d)  Hydrostatic  Arch.  —  Let  the  figure  represent  a  portion 
of  a  linear  arch  suited  to  support  a  load 
which  will  induce  in  it  a  normal  pressure  at 
every  point.  The  pressure  being  normal 
has  no  tangential  component,  and  the 
thrust  (7")  along  the  arch  must  therefore  be 
everywhere  the  same. 

Consider  any  indefinitely  small  element 
CD. 

It  is  kept  in  equilibrium  by  the  equal  -FIG'  489' 

thrusts  (7")  at  the  extremities  C  and  D,  and  by  the  pressure 
/  .  CD.  The  intensity  of  pressure  /  being  assumed  uniform 
for  the  element  CD,  the  line  of  action  of  the  pressure/.  CD 
bisects  CD  at  right  angles. 

Let   the   normals   at  C  and  D  meet  in   Ol  ,  the  centre  of 
curvature. 

Take  Of  —  O,D  =  p,  and  the  angle  CO,D  =  2  AS. 

Resolving  along  the  bisector  of  the  angle 


6  =p.  CD  —  pp 
or 

2TA&  =  pp.  2  AS  \ 
and  hence, 

T  =  pp  =  a  constant.       .     .     .     (i) 

Thus,  a  series  of  curves  may  be  obtained  in  which  p  varies 
inversely  as/,  and  the  hydrostatic  arch  is  that  curve  for  which 
\\\e  pressure  p  at  any  point  is  directly  proportional  to  the  depth 
of  the  point  below  a  given  horizontal  plane. 

Denote  the  depth  by  y,  and  let  w  be  the  specific  weight  of 
the  substance  to  which  the  pressure/  is  due.  Then 

P  =  wy,  ..........    .  .     (2) 

and 

T  —  pp  —  wyp  =  a  constant.      .     .     .     (3) 
The  curve  may  be  delineated  by  means  of  the  equation 

yp  =  const  ...........     (4) 


75  8  THEORY  OF  STRUCTURES. 

It  may  be  shown,  precisely  as  in  Case  (c),  that  the  horizontal 
intensity  of  pressure  (p^) 


=:  the  vertical  intensity  (py)  =fl (5) 

Take  as  the  origin  of  co-ordinates  the  point  O  vertically 
above  the  crown  of  the  arch,  in  the  given  horizontal  plane. 
Let  the  horizontal  line  through  O  be  the  axis  of  x. 
"       "    vertical         "  "         "    "     "       "     "  y. 

Any  portion  AM  of  the  arch  is  kept  in  equilibrium  by  the 

O equal  thrusts  (T)  at   A  and  M, 

j  and  by  the  resultant  load  P  upon 

AM,  which  must  necessarily  act 
in  a  direction  bisecting  the  angle 
ANM. 

FIG.  490.  Complete    the    parallelogram 

AM,  and  take  SN  —  NM  to  represent  T. 

The  diagonal  NL  will  therefore  represent  P. 
Let  0  be  the  inclination  of  the  tangent  at  M  to  the  hori- 
zontal. 

The  vertical  load  upon  AM  —  vertical  component  of  P 

=  LK  —  T  sin  6  =  pp  sin  0  =  wyp  sin  0  =  wy0p0  sin  0,  .     (6) 

yot  p0  being  the  values  of  y,  p,  respectively,  at  A. 

The  horizontal  load  upon  AM=  horizontal  component  of  P 


=  NK=SN-KS  =  T- 


=  2pp   (sin  -)    =  2wyp  sin  —  =  2wy0p0  (sin  -)  .     .     (7) 

\  2i'  £  •  £' 

Again,  the  vertical  load  upon  AM 
—    /  "* pdx  —  w  /    ydx  =  wy0p0  sin  0 ; (8) 

e/o  vo 

the  horizontal  load  upon  AM 

f*y  f*y  iv  /          \ 

=  J    pdy  =  wj^  ydy  =  -(/  -  y*)  =  2wy0p0  (s'm  -j  .  (9) 


EXAMPLES   OF  LINEAR   ARCHES.  759 

Equation  (8)  also  shows  that  the  area  bounded  by  the  curve 
AM,  the  verticals  through  M  and  A,  and  the  horizontal 
through  0  is  equal  to  y0pa  sin  #,  and  is  therefore  proportional 
to  sin  0.  At  the  points  defined  by  d  =  90°  the  tangents  to 
the  arch  are  vertical,  and  the  portion  of  the  arch  between  these 
tangents  is  alone  available  for  supporting  a  load.  The  vertical 
and  horizontal  loads  upon  one-half  the  arch  are  each  equal  to 

WJW 

Corollary.  —  The  relation  given  in  eq.  (i)  holds  true  in  any 
arch  for  elements  upon  which  the  pressure  is  wholly  normal. 

This  has  been  already  proved  for  the  parabola  and  catenary, 
in  cases  (a)  and  (b). 

At  the  point  A'  of  the  elliptic  arch, 

_  OB'*  _  c*r*  _ 
='-~r~~    ~ 


Hence,  the  horizontal  thrust  at  A' 

=  PyP  =  ~P  =  PCr  =  °H- 

(e)  Geostatic  Arch.  —  The  geostatic  is  a  parallel  projection  of 
the  hydrostatic  arch. 

The  vertical  forces  and  the   lengths   of  vertical  lines  are 
unchanged. 

The  horizontal  forces  and  lengths  of  hori- 
zontal lines  are  changed  in  a  given  ratio 
c  to  I. 

Let  B'  A  be  the  half-geostatic  curve  de-  FIG.  491. 

rived  from  the  half-hydrostatic  curve  BA. 

The  vertical  load  on  AB' 

—  Pf  =  P=  thrust  along  arch  at  B'.      ...     (i) 

The  horizontal  load  on  AB' 

=  H'  =  cH  —  thrust  along  arch  at  A.  .    .     .    (2) 
The  new  vertical  intensity 

-^'-—    -*-A2  , 

f>       OB' 


THEORY  OF  STRUCTURES. 

The  new  horizontal  intensity 

H1      cH 

~=C^  =  ^     ....     (4) 


Thus,  the  geostatic  arch  is  suited  to  support  a  load  so  dis- 
tributed as  to  produce  at  any  point  a  pair  of  conjugate  press- 
ures ;  pressures,  in  fact,  similar  to  those  developed  according 
to  the  theory  of  earthwork. 

Let  Rl  ,  R^  be  the  radii  of  curvature  of  the  geostatic  arch 
at  the  points  A,  B'  ,  respectively,  and  let  rlt  ry  be  the  radii  of 
curvature  at  the  corresponding  points  A,  B  of  the  hydrostatic 
arch. 

The  load  is  wholly  normal  at  A  and  B  '  .     Thus, 

H'  =pyfRl=^Rl=:cH=cprl.     ...     (5) 

•'•  R*  =  ^  .......     .....     (6) 

Also, 


•••  cR,  =  r,.  .     .....    '.     .     .     .     .     .     (8) 

(/)  General  Case.  —  Let  the  figure  represent  any  linear 
p  arch  suited  to  support  a  load  which  is  sym- 
metrically distributed  with  respect  to  the 
crown  A,  and  which  produces  at  every  point 
of  the  arch  a  pair  of  conjugate  pressures, 
the  one  horizontal  and  the  other  vertical. 

Take  as  the  axis  of  y  the  vertical  through 
the  crown,  and  as  the  axis  of  x  the   hori- 
FIG.  492.  zontal  through  an  origin   O  at  a  given  dis- 

tance from  A. 

Any  portion  A  M  of  the  arch  is  kept  in  equilibrium  by  the 
horizontal  thrust  H  at  A,  the  tangential  thrust  T  at  M,  and 
the  resultant  load  upon  AM,  which  must  necessarily  act  through 
the  point  of  intersection  N  of  the  lines  of  action  of  //and  T. 
Since  the  load  at  A  is  wholly  vertical,  H  is  given  by 

X.=P,P.,  -    ......     (i) 


EXAMPLES   OF  LINEAR  ARCHES.  jl 

p0  and  p0  being,  respectively,  the  vertical  intensity  of  pressure 
and  the  radius  of  curvature  at  A. 

Let  MN  =  T,  and  take  NS  =  H0  . 

Complete  the  parallelogram  SM\  the  diagonal  NL  is  the 
resultant  load  upon  AM  "in  direction  and  magnitude. 

The  vertical  (KL)  and  the  horizontal  (KN)  projections  of 
NL  are,  therefore,  respectively,  the  vertical  and  horizontal 
loads  upon  AM. 

Denote  the  vertical  load  by  V,  the  horizontal  by//.     Then 


(2), 

and 

H=KN=SN-SK=H0~  Fcot  0,    .    .    (3) 

0  being  the  angle  between  MN  and  the  horizon. 

dV 
pyy  the  vertical  intensity  of  pressure,  —  -j—  .    .....     (4) 

px  ,  the  horizontal   intensity  of  pressure 

">  .....      (?) 


EXAMPLE.  —  A  semicircular  arch  of  radius  r,  with  a  hori- 
zontal extrados  at  a  vertical  distance  R  from  the  centre. 
The  angle  between  the  radius  to  J/and  the  vertical  =  6. 

.'.  x  —  r  sin  #,    y  =  R  —  r  cos  0.     .     .     .     (i) 

dx=r  cos  Ode,     dy  =  r  sin  OdO  .....     (2) 

py  —  wy  =  w(R  —  r  cos  0),  ......     (3) 

w  being  the  specific  weight  of  the  load.     Hence, 
V  =  wf\R  -  r  cos  0)r  cos  BdB 

I  rtt       r  sin  2#\ 

=  wr(R  sine--  •       —  -  ].    ...     (4) 


7^2  THEORY   OF   STRUCTURES. 

Equations  (3)  and  (4)  give  H ';  for 

p.  =  w(R-r), (5) 

ind  hence 

HQ  ==  wr(R  -r) (6) 

px ,  the  horizontal  intensity  of  pressure, 

d  .  (n      r  B  -  sin  0  cos  6  _\     ,  , 

=  —  -7- (Fcotff)  =  w\R — .— -a —  rcosfl).   (7) 

dy^  \         2  sin  9  / 

Rankine  gives  the  following  method  of  determining  whether 
a  linear  arch  may  be  adopted  as  the  intrados  of  a  real  arch. 
At  the  crown  a  of  a  linear  arch  ab  measure  on  the  normal  a 
length  aCj  so  that  c  may  fall  within  the  limits  required  for 
stability  (e.g.,  within  the  middle  third). 

At  c  two  equal  and  opposite  forces,  of  the  same  magnitude 
as  the  horizontal  thrust  H  at  a,  and  acting  at  right  angles  to 
ac,  may  be  introduced  without  altering  the  equilibrium. 

Thus  the  thrust  at  a  is  replaced  by  an  equal  thrust  at  c,  and 
a  right-handed  couple  of  moment  H .  ac. 

Similarly,  the  tangential  thrust  T  at  any  point  d  of  ab 
may  be  replaced  by  an  equal  and  parallel  thrust  at  e,  and  a 
couple  of  moment  T .  de. 

The  arch  will  be  stable  if  the  length  of  de,  which  is  normal 
to  ab  at  dj  is  fixed  by  the  condition  T .  de  =  H .  ac,  and  if  the 
line  which  is  the  locus  of  e  falls  within  a  certain  area  (e.g., 
within  the  middle  third  of  the  arch  ring. 

8.  Arched  Ribs  in  Iron,  Steel,  or  Timber. — In  the  fol- 
lowing articles,  the  term  arched  rib  is  applied  to  arches  con- 
structed of  iron,  steel,  or  timber.  The  coefficients  of  elasticity 
are  known  quantities  which  are  severally  found  to  lie  between 
certain  not  very  wide  limits,  and  their  values  maybe  introduced 
into  the  calculations  with  the  result  of  giving  to  them  greater 
accuracy.  There  are  other  considerations,  however,  involved 
in  the  problem  of  the  stability  of  arched  ribs  which  still  render 
its  solution  more  or  less  indeterminate. 

It  has  been  shown  that  the  curve  of  pressure,  or  linear  arch, 


ARCHED  RIB    UNDER  A     VERTICAL   LOAD.  763 

is  a  funicular  polygon  of  the  extraneous  forces  which  act  upon 
the  real  arch.  It  is,  therefore,  also  the  b ending-moment  curve, 
drawn  to  a  definite  scale,  for  a  similarly  loaded  horizontal 
girder  of  the  same  span,  whose  axis  is  the  springing  line. 

When  the  arched  rib  carries  a  given  symmetrically  dis- 
tributed load,  it  will  be  assumed  that  the  linear  arch  coincides 
with  the  axis  of  the  rib,  and  that  the  thrust  at  any  normal 
cross-section  is  axial  and  uniformly  distributed. 

The  total  stress  at  any  point  is  made  up  of  a  number  of 
subsidiary  stresses,  of  which  the  most  important  are  :  (i)  a 
direct  thrust ;  (2)  a  stress  due  to  flexure ;  (3)  a  stress  due  to  a 
change  of  temperature.  Each  of  these  may  be  investigated 
separately,  and  the  results  superposed. 

9.  Bending  Moment  (M)  and  Thrust  (T)  at  any  Point 
of  an  Arched  Rib  under  a  Vertical  Load. — Let  ABC  be  the 
axis  of  the  rib. 

Let  D  and  E  be  points  on  the  same  vertical  line,  E  being 
D-D 


FIG.  493. 

on  the  axis  of  the  rib  and  D  on  the  linear  arch  for  any  given 
distribution  of  load. 

Resolve  the  reaction  at  A  into  its  vertical  and  horizontal 
components,  and  denote  the  latter  by  H. 

Since  all  the  forces,  excepting  H,  are  vertical,  the  difference 
between  the  moments  at  D  and  E  =  H  .  DE. 

But  moment  at  D  —  o.     Hence, 

moment  at  E  =  M=  H  .  DE. 

Let  the  normal  at  E  meet  the  linear  arch  in  D'.  Then,  if 
T  is  the  thrust  along  the  axis  at  E, 

nr  E 

Tcos  DED'  =  ff=  7>  approximately, 


or 

H  .DE  =  T.D'E  =  M. 


764  THEORY  OF  STRUCTURES. 

10.   Rib  with    Hinged   Ends  ;   Invariability  of  Span. — 

Let  ABC  be  the  axis  of  a  rib  supported  at  the  ends  on  pins  or 


FIG.  494. 

on  cylindrical  bearings.  The  resultant  thrusts  at  A  and  C 
must  necessarily  pass  through  the  centres  of  rotation.  The 
vertical  components  of  the  thrusts  are  equal  to  the  corre- 
sponding reactions  at  the  ends  of  a  girder  of  the  same  span 
and  similarly  loaded,  and  H  is  given  by  the  last  equation  in 
the  preceding  article  when  DE  has  been  found. 

Let  ADC  be  the  linear  arch  for  any  arbitrary  distribution 
of  the  load,  and  let  it  intersect  the  axis  of  the  rib  at  S.  The 
curvature  of  the  more  heavily  loaded  portion  AES  will  be 
flattened,  while  that  of  the  remainder  will  be  sharpened. 

The  bending  moment  at  any  point  E  of  the  axis  tends  to 
change  the  inclination  of  the  rib  at  that  point. 

Let  the  vertical  through  E  intersect  the  linear  arch  in  D 
and  the  horizontal  through  A  in  F. 

Let  8  be  the  inclination  of  the  tangent  at  E  to  the  hori- 
zontal. 

Let  /be  the  moment  of  inertia  of  the  section  of  the  rib 
at  £. 

Let  ds  be  an  element  of  the  axis  at  E. 

_,  Mds      H.DE.ds 

Change  of  inclination  at  E  —  dv  =  —^  =  -  -=j  --  . 

If  this  change  of  curvature  were  effected  by  causing  the 
whole  curve  on  the  left  of  E  to  turn  about  E  through  an  angle 
dO,  the  horizontal  displacement  of  A  would  be 


ARCHED  RIB    WITH  HINGED   ENDS. 

This  is  evidently  equal  to  the  horizontal  displacement  of 
£,  and  the  algebraic  sum  of  the  horizontal  displacements  of  all 
points  along  the  axis  is 

H.DE.  EF.  ds        rH.  DE  .  EF.  ds 
2-      —m-      -=J-        — -      -  =  Q,    .     .     (i) 

since  the  length  AC  is  assumed  to  be  invariable. 

Thus,  the  actual  linear  arch  must  fulfil  the  condition  ex- 
pressed by  eq.  (i),  which  may  be  written 

rDE.EF.ds 
J-       —-      -=0,         (2) 

since  H  and  E  are  constant. 

If  the  rib  is  of  uniform  section, /is  also  constant,  and  eq.  (2) 
becomes 

CDE.EF.ds=o (3) 

Also,  since  DE  is  the  difference  between  DF  and  EF, 

f(DF  ~  EF)EF.  ds=o  =f^F.  EF.  ds-J*EF*ds  (4) 

Remark. — Eq.  i  expresses  the  fact  that  the  span  remains 
invariable  when  a  series  of  bending  moments,  H .  DE,  act  at 
points  along  the  rib.  These,  however,  are  accompanied  by  a 
thrust  along  the  arch,  and  the  axis  of  the  rib  varies  in  length 
with  the  variation  of  thrust. 

Let  H0  be  the  horizontal  thrust  for  that  symmetrical  loading 
which  makes  the  linear  arch  coincide  with  the  axis  of  the  rib. 

Let  T0  be  the  corresponding  thrust  along  the  rib  at  E. 

The  shortening  of  the  element  ds  at  E  of  unit  section 

T~  T 


E 


EXAMPLE  I.  Let  the  axis  of  a  rib  of  uniform  section  and 
hinged  at  both  ends  be  a  semicircle  of  radius  r. 

Let  a  single  weight  W  be  placed  at  a  point  upon  the  rib 
whose  horizontal  distance  from  (9,  the  centre  of  the  span,  is  a. 


766 


THEORY   OF  STRUCTURES. 


The  "  linear  arch  "  (or  bending-moment  curve)  consists  of 
two  straight  lines  DA,  DC. 


FIG.  495. 

Draw  any  vertical  line  intersecting  the  axis,  the  linear 
arch,  and  the  springing  line  AC  in  E' ,  D',  F',  respectively. 

Let  OF'  =  x,  and  let  dx  be  the  horizontal  projection  upon 
AC  of  the  element  ds  at  E'. 

Then 

-^  =  cosec  E'OF'  =  -=£= , 
dx  E'F  ' 


or 


(i) 


Applying  condition  (4), 

f  D'F'rdx  +  f  D'F  'rdx  =  f  E'F'rdx, 


or 


f  D'F'dx  +  f  D'F'dx  =  f  E'F'dx, 

or  area  of  triangle  ADC  —  area  of  semicircle. 

And  if  z  be  the  vertical  distance  of  D  from  AC, 


zr  = 


ARCHED   RIB    WITH  HINGED  ENDS. 


767 


or 


nr 


z  =  —  =  one-half  of  length  of  rib. 


(2) 


nr 


(3) 


Hence,  if  h  be  the  horizontal  thrust  on  the  arch  due  to  W, 


=  M  =  W 


r*  —  a' 

2r 


(4) 


Similarly,  if  there  are  a  number  of  weights  W^  W^  PF3, .  .  , 
upon  the  rib,  and  if  h^  h^,  h^,  .  .  .  are  the  corresponding  hori- 
zontal thrusts,  the  total  horizontal  thrust  //will  be  the  sum  of 
these  separate  thrusts,  i.e., 


(5) 


It  will  be  observed  that  the  apices  (Dlt  D^,  Ds,  .  .  .)  of  the 
several  linear  arches  (triangles)  lie  in  a  horizontal  line  at  the 

nr 

vertical  distance  —  from  the  springing  line. 

Ex.  2.  An  arched  rib  hinged  at  the  ends  and  loaded  with 
weights  W,,  W^  W^  .  .  . 


-. L__      _J i._.J__ 


\ 


FIG.  497. 


Let  i  2  3  4  ...»  be  the  line  of  loads,  W^  being  represented 
by  i  2,  JF2  by  2  3,  W3   by  3  4,  etc.,  and  let  the  segments   \x, 


THEORY.  OF  STRUCTURES. 

iix,  respectively,  represent  the  vertical  reactions  at  A  and  C. 
Take  the  horizontal  length  xP  to  represent  H,  and  draw  the 
radial  lines  Pi,  P2,  P$t  .  .  . 

The  equilibrium  polygon  Ag^g^  .  .  .  must  be  the  funicu- 
lar polygon  of  the  forces  with  respect  to  the  pole  P,  and  there- 
fore the  directions  of  the  resultant  thrusts  from  A  to  £lt  El  to 
£v,  £9  to  E3,  .  .  .  are  respectively  parallel  to  Pi,  P2,  ^3,  ... 

The  tangential  (axial)  thrust  and  shear  at  any  point  p  of 
the  rib,  e.g.,  between  Et  and  Es ,  may  be  easily  found  by  draw- 
ing Pt  parallel  to  the  tangent  at/,  and  3^  perpendicular  to  PL 
The  direct  tangential  thrust  is  evidently  represented  by  Pt, 
and  the  normal  shear  at  the  same  point  by  3/.  The  latter  is 
home  by  the  web. 

If/  is  a  point  at  which  a  weight  is  concentrated,  e.g.,  £t , 
draw  Pt't"  parallel  to  the  tangent  at  £,  and  5/',  6t"  perpen- 
dicular to  Pt't". 

Pt'  represents  the  axial  thrust  immediately  on  the  left  of 
E+ ,  and  5/'  the  corresponding  normal  shear,  while  Pt"  repre- 
sents the  axial  thrust  immediately  on  the  right  of  Et,  and  6t" 
the  corresponding  normal  shear. 

A  vertical  line  through  P  can  only  meet  the  line  of  loads 
at  infinity. 

Thus,  it  would  require  the  loads  at  A  and  C  to  be  infinitely 
great  in  order  that  the  thrusts  at  these  points  might  be  vertical. 
Practically,  no  linear  arch  will  even  approximately  coincide 
•with  the  axis  of  a  rib  rising  vertically  at  the  springings,  and 
lience  neither  a  semicircular  nor  a  semi-elliptical  axis  is  to  be 
recommended. 

Ex.  3.  Let  the  axis  of  the  rib  be  a  circular  arc  of  span  21 
and  radius  r,  subtending  an  angle  2ot  at  the  centre  N. 

Let  the  angles  between  the  radii  NE,  NE'  and  the  vertical 
be  ft  and  0,  respectively. 

The  element  ds  at  E'  =  rd6. 

Also,  E'F'  =  r(cos  6  —  cos  a) ;     AF  =  /  —  r  sin  6  ; 


D'F'  =  (l ~  r  sin 


ARCHED    RIB    WITH  HINGED   ENDS.  769 


Applying  condition  (5), 

/   r»(cos  8  -  cos 'a)*rdO 

—  I     rr~(t—  r  sin  6>)r(cos  6  —  cos  a)rdB 

+    f    j^—  (l—r  sin  6>)r(cos  6  —  cos  a)rdO, 
Jtl~-a 

which  easily  reduces  to 

r\a(cos  20.  -f-  2)  —  f  sin  2#J 

—  T» — »  -j  /a(sin  ^  —  ^  cos  «)  -| —  (cos  2<af  —  cos  2ft) 

-  a    (  4 

—  rl  cos  <*(cos  <*  —  cos  ft)  —  la(sin  ft  —  ft  cos  a)  v  , 

an  equation  giving  z  or  Z?/''.     Also, 

iD^«r^/r-^£?r 

and   the  corresponding   horizontal   thrust   may  be   found,  as 
before,  by  the  equation 

/a  —  <22 


77°  THEORY  OF  STRUCTURES. 

Note—\i  a°  —  c/T, 

7t           2z     IF  —  a\  nr 

r—  --  -p—_ — ~\ />     or    z  —  —  as  in  Ex.  I. 

Ex.  4.     Let  the  axis  be  a  parabola  of  span  2/  and  rise 
(Fig.  498,  Ex.  3).     From  the  properties  of  the  parabola, 


l±a  ' 

and 

ds*  =  d 


or,  approximately, 


ds  —  dx\\       2X' 
Applying  condition  (5), 


which  easily  reduces  to 


an  equation  giving  z  or  DF. 

Note.  —  If  the  arch  is  very  flat,  so  that  ds  may  be  considered 


ARCHED   RIB    WITH  ENDS  ABSOLUTELY  FIXED.       771 


as  approximately  equal   to  dx,   the   term   2j-,^2  in   the  above 
equation  may  be  disregarded,  and  it  may  be  easily  shown  that 

16 


or 


2  = 


32    k 


II.  Rib  with  Ends  absolutely  Fixed.— Let  ABC  be  the 
axis  of  the  rib.     The  fixture  of  the  ends  introduces  two  un- 


'K  L' 

FIG.  499. 

known  moments  at  these  points,  and  since  H  is  also  unknown, 
three  conditions  must  be  satisfied  before  the  strength  of  the 
rib  can  be  calculated. 

Represent  the  linear  arch  by  the  dotted  lines  KL ;  the 
points  K,  L  may  fall  above  or  below  the  points  A,  C. 

Let  a  vertical  line  DEF  intersect  the  linear  arch  in  D,  the 
axis  pf  the  rib  in  E,  and  the  horizontal  through  A  in  F. 

As  in  Art.  10,  change  of   inclination  at  E,  or  dO,  =  —^-r. 
\  El 

But  the  total  change  of  inclination  of  the  rib  between  A  and 
C  must  be  nil,  as  the  ends  are  fixed. 


*Mds 


Mas 

-£7  =  °  = 


H.DE.ds 


which  may  be  written 


(I) 


(2) 


since  H  and  E  are  constant. 


772  THE  OR  Y  OF  STRUCTURES. 

If  the  section  of  the  rib  is  uniform,  /  is  constant  and  eq. 
(2)  becomes 


Again,  the  total  horizontal  displacement  between  A  and  C 
will  be  nil  if  the  abutments  are  immovable.  If  they  yield,  the 
amount  of  the  yielding  must  be  determined  in  each  case,  and 
may  be  denoted  by  an  expression  of  the  form  yw//,  yu  being 
some  coefficient. 

As  in  Art.  10,  the  total  horizontal  displacement 


p 
~  J 


H.DE.EF.ds 


•H.DE.EF.ds 
~ET 

But  H  and  E  are  constant. 

*DE .  EF.  ds 


(5) 


If  the  section  of  the  rib  is  uniform,  /  is  also  constant,  and 
hence 

fDE.EF.ds  =  o    or    =—;...    (6) 

and  since  DE  is  the  difference  between  DF  and  EF,  this  last 
may  be  written 


ds  =  o    or     =      .      ...     (7) 

Again,  the  total  vertical  displacement  between  A  and  C 
must  be  nil. 

The  vertical  displacement  of  E  (see  Art.  10) 


ARCHED   RIB    WITH  ENDS  ABSOLUTELY  FIXED. 
Hence,  the  total  vertical  displacement 
*H.DE.AF  , 


=  r 


which  may  be  written 

.AF 


/JJtL  .  Sif    ,  ,  . 

-f" *=° •   fe> 

since  H  and  £  are  constant.     If  the  section  of  the  rib  is  also 
constant, 


.AF.ds.     (10) 


Eqs.  (2),  (5),  and  (9)  are  the  three  equations  of  condition. 

In  eq.  (9)  AF  must  be  measured  from  same  abutment 
throughout  the  summation. 

The  integration  extends  from  A  to  C. 

EXAMPLE  I.  Let  the  axis  of  the  rib  be  a  circular  arc  of 
span  2/,  subtending  an  angle  2a  at  the  centre  N. 

Let  a  weight  W  be  concentrated  on  the  rib  at  a  point  E 


whose  horizontal  distance  from  the  middle  point  of  the  span 
is  a. 

Let  the  radius  NE  make  an  angle  /3  with  the  vertical. 

The  "  linear  arch  "  consists  of  two  straight  lines  DA',  DC' . 

Let  A  A    =  r,,  DF  =  z,CC'  =  y, . 


774  THEORY   OF  STRUCTURES. 

Draw  any  ordinate  E'F'  intersecting  the  linear  arch  in  Df . 
Let  the  radius  NE'  make  an  angle  0  with  the  vertical. 
Then 

E'F'  —  r(cos  6  —  cos  «). 
AF'  =  l—  rsintf,    and     D'F'=(l—rs\n 
if  Ff  is  on  the  left  of  F\ 

and     D'F'  =  (l-rsm 


if  /*"'  is  on  the  right  of  F. 
Also,  ^y  =  rdO. 
Applying  condition  (i), 


os  6>  —  cos  *)<#  ......     (i) 

Applying  condition  (3),  and  assuming  ^  =  o, 

/B(cos  ^  -  cos  «)  |  (/  -  r  sin  ^q^'  +^  1  ^ 
3 

+    Acos  0  -  cos  a)  I  (/  —  r  sin  ^)y^  +7,  I 
os  6>  -  cos  «r)V0  .....     (2) 


ARCHED   RIB    WITH  ENDS  ABSOLUTELY  FIXED. 
Applying  condition  (5), 


+   A/+r  sin  0)  j  (/-  rsin  tff^+f 
=  rC  a(cos  #  —  cos  «)(/  —  r  sin  0)</0 

s  0  —  cos  «)(/+  r  sin  0)</0.     (3) 


Equations  (i),  (2),  (3)  may  be  easily  integrated,  and  the  re- 
sulting equations  will  give  the  values  of  7,  ,  z,  and  yt. 

The  corresponding  horizontal  thrust,  //,  may  now  be  ob- 
tained from  the  equation  h  .  DE  =  M  =  h(z  —  EF). 

Note.  —  If  the  axis  is  a  semicircle,  and  if  Wis  at  the  crown, 

a  =  o,     a  =  90°,     ft  =  O, 
and  eqs.  (i),  (2),  (3)  reduce  to 


it  —  2  4+  2n  —  TT* 

/.  z  =  r  --  ,    and    y.  =  yt  =  r  ---  . 
4  —  TT'  4—  TT 

Ex.  2.    Let  the  axis  be  a  parabola  of  span  2/  and  rise  k 
(Fig.  500  in  Ex.  i). 

As  in  Ex.  3,  Art.  10, 


THEORY  OF  STRUCTURES. 

Also, 

D'F'  =  yl  +  (/  —  *)TTT^-  on  the  right  of  DF, 
and 

&  _.---.    <\t 

=  yt  +  (/  —  *}-7 — r?  on  the  left  of  DF. 
J*   '  '/—  ^ 


The  equations  of  condition  become 


{  ^  +  c  -  -fi~  }  0  -  ?)*  0  +  £ 


—  / 


r  j  7,  +  (/  -  *)^5  }(/-*>  (i  +  2 

+    r  j^  +  (/  -  ^-yf~  }(/+*)(>+  2 
«y  « 

r  *(i  -  ^)  (/  -  *)  (i  +  2^)^ 

«/-a 

//    /         rs 
4*  -  r 
. 


EFFECT   OF  A    CHANGE    OF'  TEMPERA  TURE.  777 

These  equations  may  be  at  once  integrated,  and  the  result- 
ing equations  will  give  the  values  of  ylt  jj>2 ,  z. 

If  the  arch  is  very  flat,  so  that  ds  may  be  taken  to  be  ap- 
proximately the  same  as  dx,  it  may  be  easily  shown  that 

2  /+  50  2    I  —  5#  6 


12.  Effect  of  a  Change  of  Temperature. — The  variation 
in  the  span  2/  of  an  arch  for  a  change  of  t°  from  the  mean 
temperature  is  approximately  =  2etl,  e  being  the  coefficient  of 
expansion. 

Hence,  if  Ht  is  the  horizontal  force  induced  by  a  change  of 
temperature,  the  condition  that  the  length  AC  is  invariable  is 
expressed  by  the  equation 

DE.EF.ds 

_£_  2eti  _  a 

If  the  rib  is  of  uniform  section,  7  is  constant;  and  since 
E  is  also  constant,  the  equation  may  be  written 


.  EF.  ds  ±  2etl  =  o. 
EXAMPLE  I.  Let  the  axis  AEC  of  a  rib  of  uniform  section 


FIG. 


be  the  arc  of  a  circle  of  radius  r  subtending  an  angle  2a  at  the 

centre. 

First,  let  the  rib  be  hinged  at  both  ends. 


7/8  THEORY  OF  STRUCTURES. 

It  is  evident  that  the  straight  line  AC  is  the  "linear  arch.'* 
Then, 

J*DE .  EF.  ds  ^j^EF^ds  =  r*  J*\cos  B  -  cos  afdB 

—  r*{a(2  -f-  cos  201)  —  |  sin  2a\. 
Also,  /  —  r  sin  a. 

H       /3 


Note. — If  the  axis  is  a  semicircle,  a  =  90°,  and 

!L?L 

EI     2 


±  2etl  =  o. 


Second,  let  the  rib  \>z  fixed  zk  both  ends. 
The  "  linear  arch"  is  now  a  straight  line  A'C'  at  a  distance 
=  Z?/*')  from  ^4  C  given  by  the  equation 

CDE.ds^O. 

.-.  CDF.  ds  =f£F.  dsy 

or 

z  Cds  =  r*  /"(cos  ^  —  cos  a)dO, 

or 

a.3-  =  r(sin  a  —  a  cos  a). 
Also, 


.  EF.  ds  =f(DF  .  EF^EF^ds  =  zj*EFds  ~  C  EF  *ds 
=  2.3T2(sin  a  —  a  cos  <*)~  r*{a(2-\-  cos  2«)  —  f  sin  2a\. 

TT       f  -^ 

.'.  -J£T\  2Jsr\sin  a  —  a  cos  a)  —  r*{a(2  +  cos  2a)  —  f  sin  2a\  ( 

±  2etl  =  o, 
and     /  =  r  sin  a. 


Ex.  2.  Let  the  axis  ^££7  of  a  rib  of  uniform  section  be  a 
parabola  of  span  2/  and  rise  k.     (See  Fig.  501  in  Ex.  I.) 


EFFEC7"  OF  A    CHANGE   OF    l^EMPERATURE.  779 

First,  let  the  rib  be  hinged  at  both  ends. 

The  straight  line  AC  is  the  linear  arch.     Then 


CDE  .EF.ds  =    fl£F2 


and  hence, 


Second,  let  the  rib  be  fixed  at  both  ends. 
The  linear  arch  is  the  line  A'C'  at  a  distance  z  (=  DF) 
from  A  C  given  by  the  equation 


or 


E  .ds  =  o  =   C(DF  ~  EF)ds, 
DFfds  =  fEF.  ds. 

*T(i  +  7^)^  =  A(i  -  f!)(i  +  2^>)^. 


or 

2  2 


Also, 

.EF.ds  =&F  .EF.ds-^   (* 


7^0  THEORY  OF  STRUCTURES. 

Hence, 


=  o. 


Remark. — The  coefficient  of  expansion  per  degree  of  Fah- 
renheit is  .0000062  and  .0000067  for  cast-  and  wrought-iron 
beams,  respectively.  Hence,  the  corresponding  total  expansion 
or  contraction  in  a  length  of  100  ft.,  for  a  range  of  60°  F.  from 
the  mean  temperature,  is  .0372  ft.  (=  ^/')  and  .0402  ft.  (=  £"). 

In  practice  the  actual  variation  of  length  rarely  exceeds  one- 
half  Q(  these  amounts,  which  is  chiefly  owing  to  structural  con- 
straint. 

13.  Deflection  of  an  Arched  Rib. 


FIG.  502. 

Let  the  abutments  be  immovable. 

Let  ABC  be  the  axis  of  the  rib  in  its  normal  position. 

Let  ADC  represent  the  position  of  the  axis  when  the  rib  is 
loaded. 

Let  BDF  be  the  ordinate  at  the  centre  of  the  span  ;  join 
AB,  AD. 

Then 


*  =  AD*  -  AF*  =  AB-  AF\ 

arc  AB 


But 

arc  AB  —  arc  AD       / 

~~          ~  ~E  ' 


/being  the  intensity  of  stress  due  to  the  change  in  the  length 
of  the  axis. 


/.  DF*  =  AB*i  -         -  AF*  = 


ELEMENTARY  DEFORMATION  OF  AN  ARCHED   RIB.     /8l 


AB*     2      - 


f\* 


=  BF*-  DF*  =  (BF-  DF)(BF  +  DF) 
=  2BF(BD\  approximately. 


^rj    is  also  sufficiently  small  to  be  disregarded.     Hence, 
h  I 


AB*  f      V  +  I*  f 
BD,  the  deflection,  =  -jr=  -^  —  —  7  —  -g  ,  approximately. 

14.  Elementary  Deformation  of  an  Arched  Rib. 


FIG.  503. 

The  arched  rib  represented  by  Fig.  503  springs  from  two 
abutments  and  is  under  a  vertical  load.  The  neutral  axis  PQ 
is  the  locus  of  the  centres  of  gravity  of  all  the  cross-sections,  of 
the  rib,  and  may  be  regarded  as  a  linear  arch,  to  which  the 
conditions  governing  the  equilibrium  of  the  rib  are  equally  ap- 
plicable. 

Let  A  A '  be  any  cross-section  of  the  rib.  The  segment 
AA'P  is  kept  in  equilibrium  by  the  external  forces  which  act 
upon  it,  and  by  the  molecular  action  at  A  A'. 

The  external  forces  are  reducible  to  a  single  force  at  C  and 
to  a  couple  of  which  the  moment  M  is  the  algebraic  sum  of 
the  moments  with  respect  to  C  of  all  the.  forces  on  the  right 
of  C. 

The  single  force  at  C  may  be  resolved  into  a  component  T 
along  the  neutral  axis,  and  a  component  Sin  the  plane  A  A'. 


THEORY  OF  STRUCTURES. 

The  latter  has  very  little  effect  upon  the  curvature  of  the  neu- 
tral axis,  and  may  be  disregarded  as  compared  with  M. 

Before  deformation  let  the  consecutive  cross-sections  BE1 
and  AA  meet  in  R ;  R  is  the  centre  of  curvature  of  the  arc 
CC'  of  the  neutral  axis. 

After  deformation  it  may  be  assumed  that  the  plane  A  A' 
remains  unchanged,  but  that  the  plane  BB'  takes  the  position 
B"B'".  Let  AA'  and  B"B"'  meet  in  R' ;  R  is  the  centre  of 
curvature  of  the  arc  CC'  after  deformation. 

Let  abc  be  any  layer  at  a  distance  z  from  C. 

Let  CC  =  As,  CR  =  R,  CR'  =  R'y  and  let  Aa  be  the  sec- 
tional area  of  the  layer  abc. 

By  similar  figures, 

ac  __  R'  +  z  ab  _  R+z 

~~A    —       r>/  and     ~ —       ~ —  • 

As          R  As          R 

i  i         i\ 

The  tensile  stress  in  abc 

-    A   be  As.z 

=  E  .  Aa—;  =  E  .  Aa -r— 

ab  ab 

z?     A        I  I         l  \  i 

=  jfi  .  Aa  .  x\~pT  —  jrjji  very  nearly. 

The  moment  of  this  stress  with  respect  to  C 


Hence,  the  moment  of  resistance  at  A  A  ' 


the  integral  extending  over  the  whole  of  the  section. 


ELEMENTARY  DEFORMATION  OF  AN  ARCHED   RIB.     783 

Again,  the  effect  of  the  force  T  is  to  lengthen  or  shorten 
the  element  CC',  so  that  the  plane  BB'  will  receive  a  motion 
of  translation,  but  the  position  of  -R'  is  practically  unaltered. 

Corollary  i.  Let  A  be  the  area  of  the  section  AAr. 

The  total  unit  stress  in  the  layer  abc 

T      Ms 


the  sign  being  plus  or  minus  according  as  M  acts  towards  or 
from  the  edge  of  the  rib  under  consideration. 

From  this  expression  may  be  deduced  (i)  the  position  of 
the  point  at  which  the  intensity  of  the  stress  is  a  maximum  for 
any  given  distribution  of  the  load;  (2)  the  distribution  of  the 
load  that  makes  the  intensity  an  absolute  maximum  ;  (3)  the 
value  of  the  intensity. 

Cor.  2.  Let  w  be  the  total  intensity  of  the  vertical  load  per 
horizontal  unit  of  length. 

Let  w,  be  the  portion  of  w  which  produces  only  a.  direct 
compression. 

Let  //be  the  horizontal  thrust  of  the  arch. 

Let  P  be  the  total  load  between  the  crown  and  AA'  which 
produces  compression. 

Refer  the  rib  to  the  horizontal  OX  and  the  vertical  OPY 
as  the  axes  of  x  and  y,  respectively. 

Let  x,  y  be  the  co-ordinates  of  C. 

Then  " 

P=ff~-'t     but     dP  =  w,dx. 
dx 


.......    (3) 

MW 

also,  : 


784  THEORY  OF  STRUCTURES. 

15.  General  Equations. 

Let  /  be  the  span  of  the  arch. 

Let  x,  y  be  the  co-ordinates  of  the  point  C  before  deforma- 
tion. 

Let  x\  y'  be  the  co-ordinates  of  the  point  C  after  deforma- 
tion. 

Let  0  be  the  angle  between  tangent  at  C  and  OX  before 
deformation. 

Let  0'  be  the  angle  between  tangent  at  C  and  OX  after 
deformation. 

Let  ds  be  the  length  of  the  element  CC'  before  deforma- 
tion. 

Let  ds  be  the  length  of  the  element  CC'  after  deformation. 

d(T        i  dO         i 

Effect  of  flexure.     ~jp  —  ~g>       and     ~fa   ~  R' 

Mi          i       dO'       de       dO'  -  dO 


Let  i  be  the  change  of  slope  at  C.     Then 

Mds       Mds 


dt  =  dO  —  dB'  = 


C 

-0'  =  f.+      / 


El     ~  Eldx 
**M  ds 


t0  being  the  change  of  slope  at  P,  and  a  quantity  whose  value 
has  yet  to  be  determined. 

Again,  the  general  equations  of  equilibrium  at  the  plane 
A  A'  are 


d 


for  the  portion  wl ,  Cor.  2,  Art.  14,  produces  compression  only 
and  no  shear. 


GENERAL   EQUATIONS.  785 

S9  being  the  still  undetermined  vertical  component  of  the  shear 

dv 
at  P,  and  ~  the  slope  at  P.     Also, 


y,  ;  -  *,    (8) 


J/0  being  the  still  undetermined  bending  moment  at  P. 

Equations  (5),  (6),  (7),  and  (8)  contain  the  four  undeter- 
mined constants  //,  50  ,  M0  ,  /„  . 

Let  Ml  ,  5t  ,  and  z,  be  the  values  of  M,  5,  and  z,  respectively, 
at  Q. 

Equations  of  Condition.  —  In  practice  the  ends  of  the  rib  are 
either  j&ritf?  or  free. 

If  they  are  fixed,  z'0  =  o  ;  if  they  are  free,  M0  =  o.  In  either 
case  the  number  of  undetermined  constants  reduces  to  three. 

If  the  abutments  are  immovable,  xl  —  /  =  o.  If  the  abut- 
ments yield,  x^  —  /  must  be  found  by  experiment.  Let  ;r,  —  / 
=  /<//,  >w  being  some  coefficient.  T\\t  first  equation  of  condi- 
tion is 


xl  —  l=o1     or     x,~l=^H.    ....     (9) 

Again,  Q   is   immovable  in   a  vertical  direction,  and    the 
second  equation  of  condition  is 


(10) 


Again,  if  the  end  Q  is  fixed,  il  =  o  ;  and  if  free,  Ml  =  o  ;  and 
the  third  equation  of  condition  is 

*t  =  o,     or     Ml  =o  ......     (il) 

Substituting  in  equations  (7)  and  (8)  the  values  of  the  three 
constants  as  determined  by  these  conditions,  the  shearing  force 
and  bending  moment  may  be  found  at  any  section  of  the  rib. 

Again, 

cos  Q'  =  cos  (0  —  i)  —  cos  0  +  i  s^n  ^  J 
sin  0'  =  sin  (0  —  i)  —  sin  6  —  i  cos  6. 


786  THEORY  OF  STRUCTURES. 

dx'       dx         ,dy  dy'       dy          dx 

•"•  ~j~r  —  ~r  +  *  j      and     jr  =  j   —  l~r-      •    (I2 
ds'       ds          ds  ds         ds         ds  v 

Hence,  approximately, 

d  .   ,  .dy  d  .dx 

•jrt*  —  x)  —  t  -j-     and     -j(  y'  —  y\  —  —  i  —. 
ds^  ds  ds^  ds 

Thus,  if  JTand  Fare  respectively  the  horizontal  and  verti- 
cal displacements, 

dX       .dy  dY  .dx 

-j-  =  t—r-     and     —r  =.  —  t—r, 
as          as  ds  ds 

or 

dX       .  dY 


16.  Effect  of  T  and  of  a  Change  of  t°  in  the  Temperature. 


Also,  if  there  is  a  change  from  the  mean  of  t°  in  the  tem- 
perature, the  length  ds\i  —  T^J  must  be  multiplied  by 
(i  ±  e/),  e  being  the  coefficient  of  linear  expansion. 


—  -p-j  ±  et),  approximately.          (14) 
By  equations  (i  2), 

<&  =  (dx  +  i  .  dyy^  =  (dx  +  i  .  dy)(i  -jj±  &) 
and 

dy'  =  (dy  -  i.dx)^  =  (dy  -  i.dx](i  -  ~  ±  et). 


GENERAL   EQUATIONS.  787 


.-.  dX  =  d(x'  -x)^idy-- 
and 


=  d(yf  —y)  =  —  idx  -    —  ^   etdy,  approximately, 
Hence, 


and 


Note. — A  nearer  approximation  than  is  given  by  the  pre- 
ceding results  may  be  obtained  as  follows: 

Let  x  +  dx,  y  -\-  dy  be  the  co-ordinates  of  a  point  very 
near  C  before  deformation. 

Let  x'  +  dx' ,  y'  -f-  dy'  be  the  co-ordinates  of  a  point  very 
near  C  after  deformation. 

Then 

ds*  =  dx*  +  dy*     and     ds'*  =  dx'*  +  dy'\ 

.-.  ds'*  -  ds*  =  dx'*  -  dx*  +  dy'*  -  dy'*, 
or 

(ds'-ds)(dsf  +  ds)  =  (dx'-dx)(dxf  +  dx)  +  (dy'-dy)(dy:  +  dy). 
••.  (ds'  —  ds)ds  =  (dx'  —  dx)dx  +  (dy'  —  dy)dyy  approximately. 

.  ,  ,  .ds  dy 

.'.  dx  —  dx  =  (ds  —  ds)-j-  —  (dy  —  dy)-~ 

ax  'ax 

and 

ds  dx 


dy'  -dy  =  (ds' 

Hence,  by  equations  (12)  and  (14), 

.dy  J  T(ds 

dx'  —  dx  —  t-^j-dx  —  -=r-.(  —I  dx  ±  eti--    dx 
dx  EA\dx'  ^dx 


788  THEORY  OF   STRUCTURES. 

and 

T  (ds\dx  J  lds\dx  , 

dy>  -dy=~  ^^—-dz  ±  - 


/.  ay  /™  1   i  ds\  rx  tds\ 

i  -j-dx  —     I    -rr-Ti  j—  1  ax  ±  et  I     I  -j-  1  dx 
dx  «/„    hA\dx I  J     \dxl 

0  '  0 

and 

r  IdsVdx  ,  r*tds\*dx 


These  equations  are  to  be  used  instead  of  equations  (15) 
and  (16),  the  remainder  of  the  calculations  being  computed 
precisely  as  before. 

The  following  problems  are,  in  the  main,  the  same  as  those 
given  in  Art.  180  of  Rankine's  Civil  Engineering,  I3th  edition- 

17.  Rib  of  Uniform  Stiffness.  —  Let  \hzdepth  and  sectional 
form  of  the  rib  be  uniform,  and  let  its  breadth  at  each  point 
vary  as  the  secant  of  the  inclination  of  the  tangent  at  the  point 
to  the  horizontal. 

Let  A),  /j  be  the  sectional  area  and  moment  of  inertia  at 
the  crown. 

Let  A,  I  be  the  sectional  area  and  moment  of  inertia  at  any 
point  C,  Fig.  503. 

Then 


(17) 


Also,  since  the  moments  of  inertia  of  similar  figures  vary 
as  the  breadth  and  as  the  cube  of  the  depth,  and  since  the 
depth  in  the  present  case  is  constant, 


(18) 


T       HsecO      H 
Again,  -j-  =  -  -    =  -j-,  and  the  intensity  of  the  thrust 


is  constant  throughout. 


ARCHED  RIB   OF   UNIFORM  DEPTH.  789 

Hence,  equations  (5),  (15),  and  (16),  respectively,  become 


/•* 
-*=J* 


dy 


H 


(21) 


Equation  (19)  shows  that  the  deflection  at  each  point  of  the 
rib  is  the  same  as  that  at  corresponding  points  of  a  straight 
horizontal  beam  of  a  uniform  section  equal  to  that  of  the  rib 
at  the  crown,  and  acted  upon  by  the  same  bending  moments. 

Ribs  of  uniform  stiffness  are  not  usual  in  practice,  but  the 
formulae  deduced  in  the  present  article  may  be  applied  without 
sensible  error  to  flat  segmental  ribs  of  uniform  section. 

18.  Parabolic  Rib  of  Uniform  Depth  and  Stiffness,  with 
Rolling  Load;  the  Ends  fixed  in  Direction;  the  Abut- 
ments immovable. 

D  E  O 


FIG. 


504- 


Let  the  axis  of  x  be  a  tangent  to  the  neutral  curve  at  its 
summit. 

Let  k  be  the  rise  of  the  curve. 

Let  x,  y  be  the  co-ordinates  at  any  point  C  with  respect 
to  O. 

Then 


and 


d~x-~    -T\~2-*>' 


79°  THEORY  OF  STRUCTURES. 

Let  -w  be  the  .dead  load  per  horizontal  unit  of  length. 

«    2£/  "     "     live       "       "  "  "      "       " 

Let  the  live  load  cover  a,  length  DE,  =  r/,  of  the  span. 
'Denote  by  (A)  formulae  relating  to  the  unloaded  division 
OEj  and  by  (B)  formulae  relating  to  the  loaded  division  DE. 
.Equations  (7)  and  (8),  respectively,  become 

(o  Z,  JLT  \ 

—  -™)*;     .........    (24) 

(B)     S=S,+  ^f  -  w}X  -  w'\x  -  (i  -  r)l\.     .    .    (25) 

;.     ......     (26) 


(B)    M  =  M,  +  S.x+--w-\x-(i-r)l\>.    (27) 

Since  the  ends  are  fixed, 

*'.  =0  =  *;  ......     .     .     (28) 

Hence,  by  equations  (19)  and  (26), 

I      '  (UH 


I    (  x} 

(A)          ,=  --^  ^Ms  +  S.—  +(-jr  -w)^  J;    .     (29) 

and;hy  equations  (19)  and  (27), 

i        .  IZkH 


.     (30) 


i=     - 


When  x  =  /,  i  =  *',  =  o,  and  therefore,  by  the  last  equation, 


\-  -  —  •/• 

)9-H-\/2     —W)A       &r  l*-   •     •    •    (30 


ARCHED    RIB    OF    UNIFORM   DEPTH.  791 


dv 
Again,  let  i  =  -- .     Then 


/"'  .dy  J  Cldv  dy  ,         .dy          Cl    d*y  , 

I  ?£*"=   /  Txixdx=l^x-  /   v^dx' 

*/  o  «y  o  */  o 


But  i,  =  o,     and      -       =  --. 


ri  dy  ,        s&   ri  u   ri  r*mj 

•'•  /  V1*^  ~  ~T  /  vdx=-~r  \  7  ^ 

ty  o  t/o  t/o    */o 

By  the  conditions  of  the  problem,  x'  —  x  and  j/  —7  are 
each  zero  at  Q.  Hence,  equations  (20)  and  (21),  respectively, 
become 


•  •  (33) 

=-J0    MX.      ..........     (34) 


Substitute  in  eqs.  (33)  and  (34)  the  value  of  i  given  by  eq. 
(30),  and  integrate  between  the  limits  o  and  /.     Then 

i*        r     tun      \  r  r 


o  --  h 
"        ' 


c  o  o  --         ---      --- 

EIl  (     °6  '     "24  '    \  I*  J  120  120 


and 

ss.tuff 


i  IMS 

-17.1— 

which  may  be  written 


792  THEORY   OF  STRUCTURES. 

and 


Hence,  by  eqs.  (31),  (35),  (36), 


4  '  '  3 


When  x  =  l,M=Mlt  and  5=5,. 
•  Hence,  by  eqs.  (25)  and  (27), 


and 


\/a      wWa 
—  w\ . 

/2  2 


-AjH; (37) 

:('-Z')+f^ (38) 


15£, 
H=-      ---  _-^_ *J.  .  .  .  (39) 


Substituting  in  these  equations  the  values  of  S0,  M0,  given 
above,  we  have 


,    .     .     .     (40) 

A  »  £•  I  l> 

and 


r  =  -  —  -  M//VI-  -  -r  +  -)  +  -kH.  .          (41) 
12  \2       3         4/3 

« 
To  find  the  greatest  intensity  of  stress,  etc. — The  intensity  of 

T        f-f 

the  stress  due  to  direct  compression  —  —*=-*-• 


ARCHED  RIB   OF    UNIFORM  DEPTH. 


793 


The  intensity  of  the  stress  in  the  outside  layers  of  the  rib 
due  to  bending  is  the  same  as  that  in  the  outside  layers  of  a 
horizontal  beam  of  uniform  section  A1  acted  upon  by  the  same 
moments  as  act  on  the  rib,  for  the  deflections  of  the  beam  and 
rib  are  equal  at  every  point  (eq.  (19) ).  Also,  since  the  rib  is 
fixed  at  both  ends,  the  bending  moment  due  to  that  portion  of 
the  load  which  produces  flexure  is  a  maximum  at  the  loaded 
end,  i.e.,  at  Q.  Hence,  the  maximum  intensity  of  stress  (/,) 

occurs  at  Q,  and/,  =  —r-  ±  -A/,  7,  zl  being  the  distance  of  the 
AI  /, 

layers  from  the  neutral  axis. 

H  and  Ml  are  both  functions  of  r,  and  therefore /j  is  an  ab- 
solute maximum  when 


But 


and 


dp, 

o  — 

i 

dH 

8. 

\-     ' 

dM, 

dr 
dH 

150 

~A 
//a 

dr  - 

LA 

dr  

r}* 

dr 

4 

k 

i-t- 

15. 

7,    ' 

dM. 


V44) 


Hence,  /,  is  an  absolute  maximum  when 


The  roots  of  this  equation  are 
r  =  i 


and 


2  4 

r  =  ±  7 


53     A 


1 


±  I 


(45) 


794 


THEORY  OF  STRUCTURES. 


11  . 

r  —  I  makes  -j-r-  zero,   so   that   the  maximum  value  of  p. 
dr* 


corresponds  to  one  of  the  remaining  roots. 
Thus, 


thewwwr.  thrust  —  ' 


and 


(46) 


the  max.  tension  =     -—  H+  -jMj  =  //',       (47) 


the  values  of  H  and  Ml  being  found  by  substituting  in  eqs. 
(39)  and  (40 

i+^^- 
2          4  Aft 


or 


.    ,  45 
_2_   '  4 


(48) 


according  as  the  stress  is  a  thrust  or  a  tension. 

If  eq.  (47)  gives  a  negative  result,  there  is  no  tension  at  any 
point  of  the  rib. 

Note.  —  The  moment  of  inertia  may  be  expressed  in  the  form 


q  being  a  coefficient  depending   upon  the  form  of  the  section. 
Hence, 

the  maximum  intensity  of  stress  =  — (  ±  H '-|-  — l ) .     .     (:tg) 

Corollary  I. — If  the  depth  of  the  rib  is  small  as  compared 
with  k,  the  fraction  j  will  be  a  small  quantity,  and  the  maxi- 
mum intensity  of  stress  will  approximately  correspond  to  r  =  -J-. 


ARCHED  RIB   OF   UNIFORM   STIFFNESS.  795 

The  denominator  in  eq.  (39)  may  be  taken  to  be  k,  and  it  may 
be  easily  shown  that  the  values  of  //,//'  are 


:-  (50) 


,_    i    («,/•/      i       iS*,\      5rf£/        54   v/r\ 

'  -  T,  I  TV  £  +  7^*4^1  +  3125^:  5'(5I) 


.  2.  If   the    numerator  in  eqs.  (48)  is  greater  than  the 
denominator,  then  r  must  be  unity.     Hence,  by  eq.  (39)  and 


and  by  eqs.  (38)  and  (41), 


Thus,  //,/,"  can  be  found  by  substituting  these  values  of 
H  and  Ml  in  eqs.  (46)  and  (47). 

19.  Parabolic  Rib  of  Uniform  Stiffness,  hinged  at  the 
Ends. 

Let  the  rib  be  similar  to  that  of  the  preceding  article. 

Since  the  ends  are  hinged,  M9  =  o  =  Ml  ,  while  i  is  an  un- 
determined constant. 

The  following  equations  apply  : 


(A)  S=S.+f=_-WJ:r; (54) 

(B)  S=S.+  K-w)*-v/\x-(i-rW;          (55) 


79^  THEORY  OF  STRUCTURES. 

(A)        M  =Ss  +      ?  -«;•.    .......    (56) 


(B)          M=S,x  +  -w-{x-(l-r-}l\\       (57) 

%kH 


(58) 


!-*  -         +  --  «    -   i*-(-^°  •  (59) 

Assume  that  the  horizontal  and  vertical  displacements  of 
the  loaded  end  are  nil. 

Substitute  in  eqs.  (20)  and  (21)  the  value  of  i  given  by  eq. 
(59).  Integrate  and  reduce,  neglecting  the  term  involving  the 
temperature.  Then 


_«//.J.l  _^  JL.    :  (6o) 


S6ff         Vs  ,. 

- 


From  (57),  since  M,  =  o, 


\/  ,7r3 

-^-  -wj--ze;7-.     .     .     .     (62) 

Equations  (60),  (61),  and  (62)   are  the  equations  of  condi- 
tion. 

Subtract  (61)  from  (60).     Then 

UH         \r  lr*      r*\          H 

"/r  - 

which  may  be  written 


ARCHED   RIB    OF    UNIFORM   STIFFNESS.  797 

Subtract  (63)  from  (62).     Then 


Hence, 

/'U  +    V°-5r;+2o} 

L"     •    •     (6S) 


Eliminating  50  between  (61)  and  (62), 
UH         \/3 


Also,  by  (55), 

r  -  »/  -  ^V/  =  -  />  suppose.     (67) 


Eliminating  50  between  (62)  and  (67), 

_/,=5i=(^_^_<4_.  .  .  .  (68) 

Eqs.  (62),  (65),  (66),  and  (68)  give  the  values  of  Hy  S0,Slt 
and  i0. 

Again,  the    maximum    bending  moment   M'  occurs   at  a 

dM 

point  given  by  —j-  =  o  in  (57),  i.e., 


-w'\x-(i-r)l\.      .(69) 
Subtract  (69)  from  (67).     Then 

_/>,  =  5,  =  (^?-  «;)(/-  *)  -  «/(/-*). 


79$  THEORY  OF   STRUCTURES. 

Hence,  the  distance  from  the  loaded  end  of  the  point  at 
which  the  bending  moment  is  greatest  is 


w  -- 


Substitute  this  value  of  x  in  (57),  and,  for  convenience,  put 


w  --w  —  —pr  —  m> 
Then 

p 


M,  =    /  - 

°  m 


'-m)l-  W'rl\ 


P'/w'  -  m       a 
m"\      2~      ' 


But  by  (62),  o  =  S. 


Hence,  M' ',  the  maximum  bending  moment, 

\ 

(71) 


As  before,  the  greatest  stress  (a  thrust) 

=  ±-H+M=P;,    ....   (72) 


and  the  value  of  r  which  makes  //  an  absolute  maximum  is 
given  by  ~-  =  o.     But  by  (71),  M'  involves  r10  in  the  numera- 


ARCHED  RIB   OF   UNIFORM  STIFFNESS.  799 

dpi 

tor  and  r5  in   the   denominator,  so  that  -~  =  o  will   be  an 

ar 

equation  involving  r14. 

One  of  its  roots  is  r  =  i,  which  generally  gives  a  minimum 
value  of  //.  Dividing  by  r  —  i,  the  equation  reduces  to  one 
of  the  thirteenth  order,  but  is  still  far  too  complex  for  use.  It 
is  found,  however,  that  r  =.  %  gives  a  close  approximation  to  the 
absolute  maximum  thrust. 

With  this  value  of  r,  and,  for  convenience,  putting 


15  I,    i 

"    "' 


By  (65), 
By  (62), 

By  (68), 
By  (66), 
By  (70), 


n  —  i      w' 

(74) 


-   I          W' 

(75) 


(77) 

r^TJ-^-+y 

By  (70. 


.  —  If  the  rib  is   merely  supported  at  the  ends  but  not 
fixed,  the.  horizontal  displacement  of  the  loaded  end  may  be 


800  THEORY  OF  STRUCTURES. 

represented  by  ^H  (Art.  1 1).  Thus  the  term  —  ^H  must  be 
added  to  the  right-hand  side  of  eq.  (15). 

20.  Parabolic  Rib  of  Uniform  Stiffness,  hinged  at  the 
Crown  and  also  at  the  Ends. — In  this  case  M=o  at  the 
crown,  which  introduces  a  fourth  equation  of  condition. 

By  (57). 

L    (UH _    ^_^Tf_l      V 
0  ~~  s°2  ~r"  \~r~  '  wls  ~    2  \    2  "•"  rl ' 

which  may  be  written 

>r-r        \  J  f  i\ 

'-r  +  -).    •    •    (79) 

Eliminating  S0  between  (79)  and  (62), 

UH 

——  —  w  =  w  (—  2r  -\-  ^r  —  i). 

Hence, 

H  =  -7}^  —  w'(2r*  —  4r+  i)}.     .     .     .     (80) 
8£( 

By  (79)» 

S0=*^-(3r*  -4/-+I) (81) 

By  (68), 
By  (66), 

By  (70)  and  (82), 

l-z=±J  =L.     . (84) 

By  (71), 


PARABOLIC  RIB    OF    UNIFORM  STIFFNESS.  8oi 

When  r  =  j, 


w'r  wfr  \  '  (86) 

t0  =  —  -—  -77-7-,  and          M'  =  -?—. 

384  hi,  64  J 

These  results  agree  with  those  of  (73)  to  (78),  if  n  =  i. 
In  general,  when  n  —  I, 

/ 
w-\-  —  (5ra  —  5r4  -f-  2r6)  =.  w  —  w\2r*  —  ^r  -f-  i), 

by  (65)  and  (80).     Hence, 


and  the  roots  are  r  —  j-,  r  =  i,  r  =  ±  4/2. 

Hence,  w  =  I  only  renders  the  expressions  in  (86)  identical 
with  the  corresponding  expressions  of  the  preceding  article 
when  n  =  J  or  i. 

Again,  the  intensity  of  thrust  is  greatest  at  the  outer  flange 
of  the  loaded  and  the  inner  flange  of  the  unloaded  half  of  the 
rib,  and  is 

r   uw      a    .  w1 


The  intensity  of  tension  is  greatest  at  the  inner  flange  of  the 
loaded  and  the  outer  flange  of  the  unloaded  half  of  the  rib, 
and  is 

w'        if          w' 


The  greatest  total  horizontal  thrust  occurs  when  r  =  I,  and 
its  value  is 


8O2  THEORY  OF  STRUCTURES. 

21.  Maximum  Deflection  of  an  Arched  Rib.  —  The  deflec- 
tion must  necessarily  be  a  maximum  at  a  point  given  by  i  =  o. 
Solve  for  x  and  substitute  in  (16)  to  find  the  deflection  y'  —  y; 

the  deflection  is  an  absolute  maximum  when  -j\y'  —  y)  =  o. 

The  resulting  equation  involves  r  to  a  high  power,  and  is  too 
intricate  to  be  of  use.  It  has  been  found  by  trial,  however, 
that  in  all  ordinary  cases  the  absolute  maximum  deflection 
occurs  at  the  middle  of  the  rib,  when  the  live  load  covers  its 

v/hole  length,  i.e.,  when  x  —  -,  and  r  =  I. 

CASE  I.  Rib  of  Art.  18.    For  convenience,  put  I  +  —  ~^  =  s. 
Then,  by  (39), 


By  (38)  and  (41), 

-M.=  ^  +  <V'f-^^^^  =  -M,..     (88) 
By  (36)  and  (38), 


S.=  -6      .............     (89) 

By  (30),  (38),  (89), 

i=-jTJ  \M«X  ~  3M°J  +  2M*J\    •••(9°> 
Hence,  the  maximum  deflection 

C"  .,  M,   f*l  x*    ,      x\  MJ' 

~J(  ldx  =  ~  Tit  (*  ~  i-T  +  2r)dx  =  ~  F/32 
r  w  +  w'  *  —  i      5  et  r 

=:  —  -  n:  —  7*  —  T  —  d.  ,  suppose.     .     .     (01  ) 

384       hi,          s        ~  128  s  k 


MAXIMUM  DEFLECTION   OF  AN  ARCHED  RIB.          803 

The  central  deflection  d^  of  a  uniform  straight  horizontal 
beam  of  the  same  span,  of  the  same  section  as  the  rib  at  the 
crown,  and  with  its  ends  fixed,  is 


Hence,  neglecting  the  term  involving  the  temperature, 

4  =  ^4  .........    (93) 

CASE  II.  Rib  of  Art.  19. 
By  (65), 


By  (66)  and  (62), 


By  (30),  (94),  and  (95), 


Hence,  the  maximum  deflection 


n 


^.  (97) 


If  the  ends  of  the  beam  in  Case  I  are  free,  its  central  de- 
flection 

5  *l 


El 


_ 


(98) 


Thus,  the  deflection  of  the  arched  rib  in  both  cases  is  less 
than  that  of  the  beam. 


804 


THEORY  OF  STRUCTURES. 


22.  Arched  Rib  of  Uniform  Stiffness  fixed  at  the  Ends 
and  connected  at  the  Crown  with  a  Horizontal  Distribut- 
ing Girder.  —  The  load  is  transmitted  to  the  rib  by  vertical 
struts  so  that  the  vertical  displacements  of  corresponding 
points  of  the  rib  and  girder  are  the  same.  The  horizontal 
thrust  in  the  loaded  is  not  necessarily  equal  to  that  in  the  un- 
loaded division  of  the  rib,  but  the  excess  of  the  thrust  in  the 
loaded  division  will  be  borne  by  the  distributing  girder,  if  the 
rib  and  girder  are  connected  in  such  a  manner  that  the  hori- 
zontal displacement  of  each  at  the  crown  is  the  same. 

The  formulae  of  Art.  18  are  applicable  in  the  present  case 
with  the  modification  that  /,  is  to  include  the  moment  of 
inertia  of  the  girder. 

The  maximum  thrust  and  tension  in  the  rib  are  given  by 
equations  (64)  and  (65). 

Let  z'  be  the  depth  of  the  girder,  A'  its  sectional  area. 


The  greatest  thrust  in  the  girder  =  —  —  ;  —  —  -]  ---  ~.     (99) 


MJ_ 
~2EL 


The  greatest  tension  in  the  girder  = 


2EL 


h~-  (I0°) 


H  and  Ml  being  given  by  equations  (66)  and  (67),  respectively. 

The  girder  must  have  its  ends  so  supported  as  to  be  capable 
of  transmitting  a  thrust. 

23.  Stresses  in  Spandril  Posts  and  Diagonals. — Fig.  505 
represents  an  arch  in  which  the  spandril  consists  of  a  series  of 
vertical  posts  and  diagonal  braces. 


0        1 


n     n+l 


FIG.  505. 


Let  the  axis  of  the  curved  rib  be  a  parabola.  The  arch  is 
then  equilibrated  under  a  uniformly  distributed  load,  and  the 
diagonals  will  be  only  called  into  play  under  a  passing  load. 


STRESSES  IN   SPANDRIL   POSTS  AND  DIAGONALS.     805 

Let  x,  y  be  the  co-ordinates  of  any  point  F  of  the  parabola 
with  respect  to  the  vertex  C.  Then 

4& 

y  =  TT*  • 

Let  the  tangent  at  Fmeet  CB  in  Z,  and  the  horizontal  BE 
in  G. 

Let  BC  =  k'.     Then 

BL^BC-  CL  =  BC-  CN=k'  -y. 

Let  Nbe  the  total  number  of  panels. 

Consider  any  diagonal  ED  between  the  nth  and  (n  -f-  i)th 
posts. 

Let  w   be  the  greatest  panel  live  load. 

The  greatest  compression  in  ED  occurs  when  the  passing 
load  is  concentrated  at  the  first  n  —  I  panel  points. 

Imagine  a  vertical  section  a  little  on  the  left  of  EF. 

The  portion  of  the  frame  on  the  right  of  this  section  is 
kept  in  equilibrium  by  the  reaction  R  at  P,  and  by  the  stresses 
in  the  three  members  met  by  the  secant  plane. 

Taking  moments  about  G, 

D.GE  cos  B  =  R.AG, 

D  being  the  stress  in  DE,  and  0  the  angle  DEP. 
Now, 

w'n(n  —  i) 

~2—  7v^' 

Also, 


"  k'  -  y  '  2y 
and  hence, 

^  z?       r*  D   i            k'x  -\-  xy  A     r  A       I    >   k'x  —  x 

GE  —  GB  -4-  x  =  -  •  —  --,  and     GA  =  —  --- 

2J  22^ 

Hence, 


2         A^  IT  j:  +  xy 

The  stresses  in  the  counter-braces  (shown  by  dotted  lines  in 
the  figure)  may  be  obtained  in  the  same  manner. 


806  THEORY   OF  STRUCTURES. 

The  greatest  thrust  in  EF  ~  w'  -|-  w. 

The  greatest  tension  in  EF  =  ZJcos  0  —  w,  w  being  the 
dead  load  upon  EF. 

If  the  last  expression  is  negative,  EF  is  never  in  tension. 

24.  Clerk  Maxwell's  Method  of  determining  the  Re- 
sultant Thrusts  at  the  Supports  of  a  Framed  Arch.  —  Let 
As  be  the  change  in  the  length  s  of  any  member  of  the  frame 
under  the  action  of  a  force  P,  and  let  a  be  the  sectional  area  of 
the  member.  Then 


. 

Ea 

the  sign  depending  upon  the  character  of  the  stress. 

Assume  that  all  the  members  except  the  one  under  con- 
sideration are  perfectly  rigid,  and  let  Al  be  the  alteration  in 

the  span  /  corresponding  to  As.      The  ratio  —  is  equal  to  a 

constant  m,  which  depends  only  upon  the  geometrical  form  of 
the  frame. 

.•.  Al  =  m  .  As  —  ±  mP-^-  . 

Ea 

Again,  P  may  be  supposed  to  consist  of  two  parts,  viz.,/, 
due  to  a  horizontal  force  H  between  the  springings,  and  /2  due 
to  a  vertical  force  V  applied  at  one  springing,  while  the  other 
is  firmly  secured  to  keep  the  frame  from  turning. 

By  the  principle  of  virtual  velocities, 

/,    M 


Similarly,  -£  is  equal  to  some  constant  n,  which  depends 
only  upon  the  form  of  the  frame. 


=-  ±  (m*H+-  mnV\~. 


CLERK  MAXWELL'S  METHOD. 
Hence,  the  total  change  in  /  for  all  the  members  is 


If  the  abutments  yield,   let  ^Al  =  }*H,  ^  being  some  co- 
efficient to  be  determined  by  experiment.     Then 


If  the  abutments  are  immovable,  241  is  zero,  and 

(D) 


2  (m*~ 


the  same  as  the  corresponding  reaction  at  the  end  of  a 
girder  of  the  same  span  and  similarly  loaded.  The  required 
thrust  is  the  resultant  of  H  and  F,  and  the  stress  in  each 
member  may  be  computed  graphically  or  by  the  method  of 
moments.  In  any  particular  case  proceed  as  follows  : 

(1)  Prepare  tables  of  the  values  of  m  and  n  for  each  member. 

(2)  Assume  a  cross-section  for  each  member,  based  on  a 
probable  assumed  value  for  the  resultant  of  V  and  H. 

(3)  Prepare  a  table  of  the  value  of  w?--  for  each  member, 


and  form  the  sum  '2{m'i^~ 
\     Ea 

(4)  Determine,  separately,  the  horizontal  thrust  between 
the  springings  due  to  the  loads  at  the  different  joints.  Thus, 
let  vl ,  v^  be  the  vertical  reactions  at  the  right  and  left  supports 


due  to  any  one  of  these  loads.     Form  the  sum 

using  vl  for  all  the  members  on  the  right  of  the  load  and  v^  for 
all  those  on  its  left.     The  corresponding  thrust  may  then  be 


808  THEORY   OF  STRUCTURES. 

found  by  eq.  (C)  or  eq.  (D),  and  the  total  thrust  H  is  the  sum 
of  the  thrusts  due  to  all  the  weights  taken  separately. 

(5)  Repeat   the  process   for  each  combination  of  live  and 
dead  load  so  as  to  find  the  maximum  stresses  to  which  any 
member  may  be  subjected. 

(6)  If  the  assumed  cross-sections  are  not  suited  to  thes^ 
maximum    stresses,  make   fresh   assumptions   and   repeat   the 
whole  calculation. 

The  same  method  may  be  applied  to  determine  the  result- 
ant tensions  at  the  supports  of  a  framed  suspension-bridge. 


Note. — The  formulae  for  a  parabolic  rib  may  be  applied 
without  material  error  to  a  rib  in  the  form  of  a  segment  of  a 
circle.  More  exact  formulae  may  be  obtained  for  the  latter  in 
a  manner  precisely  similar  to  that  described  in  Arts.  18-22, 
but  the  integrations  will  be  much  simplified  by  using  polar  co- 
ordinates, the  centre  of  the  circle  being  the  pole. 


EXAMPLES.  809 


EXAMPLES. 

1.  Assuming  that  an  arch  may  be  divided  into  elementary  portions 
by  imaginary  joint  planes  parallel  to  the  direction  of  the  load  upon  the 
arch,  find  the  limiting  span  of  an  arch  with  a  horizontal  upper  surface 
and  a  parabolic  soffit  (latus  rectum  =  40  ft.),  the  depth  over  the  crown 
being  6  ft.  and  the  specific  weight  of  the  load  120  Ibs.  per  cubic  foot; 
the  thrust  at  the  crown  is  horizontal  (=  P)  and  4  ft.  above  the  soffit. 

2.  A  masonry  arch  of  90  ft.  span  and  30  ft.  rise,  with  a  parabolic  in- 
trados  and  a  horizontal  extrados,  springs  from  abutments  with  vertical 
faces   and  10  ft.  thick,  the  outside  faces  being  carried  up  to  meet  the 
extrados.     The  depth  of  the  keystone  is  3  ft.     The  centre  of  resistance 
at  the  springing  is  the  middle  of  the  joint,  and  at  the  crown  12  in.  below 
the  extrados.     The  specific  weight  of  the  masonry  may  be  taken  at  150 
Ibs.  per  cubic  foot.     Determine  (a)  the  resultant  pressure  in  the  vertical 
joint  at  the  crown  ;  (&}  the  resultant  pressure  in  the  horizontal  joint  at 
the  springing ;  (c)  the  maximum  stress  in  the  vertical  joint  aligning  with 
the  inside  of  an  abutment. 

3.  The  intrados  of  an  arch  of  100  ft,  span  and  20  ft.  rise  is  the  segment 
of  a  circle.     The  arch  ring  has  a  uniform  thickness  of  3  ft.  and  weighs 
140  Ibs.  per  cubic  foot ;  the  superincumbent  load  may  be  taken  at  480 
Ibs.  per  lineal  foot  of  the  ring.     Determine  the  mutual  pressures  at  the 
key  and  springing,  their  points  of  application  being  2  ft.  and  i£  ft.,  re- 
spectively, from  the  intrados.     Also  find  the  curve  of  the  centres  of  pres- 
sure. 

4.  The  soffit  of  an  arch  of  30  ft.  span  and  10  ft.  rise  is  a  transformed 
catenary.     The  masonry  rises   10  ft.  over  the  crown,  and   the   specific 
weight  of  the  load  upon  the  arch  may  be  taken  at  120  Ibs.  per  cubic  foot. 
Determine  the  direction  and  amount  of  the  thrust  at  the  springing. 

5.  A  concrete  arch  has  a  clear  spring  of  75  ft.  and  a  rise  of  7^  ft. ;  the 
height  of  masonry  over  crown  =  5  ft. ;  the  weight  of  the  concrete  =  144 
Ibs.  per  cubic  foot.     Determine  the  transformed  catenary,  the  amount 
and  direction  of  the  thrust  at  the  springing,  and  the   curvatures  at  the 
crown  and  springing. 

Ans.  m  —  23.9  ;  thrust  =  91,354  Ibs. ;  slope  at  springing  =  25!° ; 
radius  of  curvature  =  114.2  ft.  at  crown  and  =248.7  ft.  at 
springing. 

6.  Determine  the  transformed  catenary  for  an  arch  of  60  ft.  span  and 
15  ft.  rise,  the  masonry  rising  6  ft.  over  the  crown  and  weighing  120  Ibs. 
per  cubic  foot.     Also  find  the  amount  and  direction  of  the  thrust  at  the 
abutments. 


8 10  THEORY  OF  STRUCTURES. 

7.  Determine  the  transformed  catenary  for  an  arch  of  30  ft.  span  and 
7^  ft.  rise,  the  height  of  masonry  over  the  crown  being  4^  ft.  ;  weight  of 
the  masonry  =  125  Ibs.  per  cubic  foot.    Also  find  the  thrust  at  the  spring- 
ing and  the  curvature  at  the  crown  and  the  springing. 

8.  In  a  parabolic  arch  of  50  ft.  span  and  10  ft.  rise,  hinged  at  both 
ends,  a  weight  of  i  ton  is  concentrated  at  a  point  whose  horizontal  dis- 
tance from  the  crown  is  10  ft.     Find  the  total  thrust  along  the  axis  oi 
tUe  rib  on  each  side  of  the  given  point,  allowing  for  a  change  of  60°  from 
the  mean  temperature  (e  =  .0000694). 

9.  A  parabolic  arched  rib  of  100  ft.  span  and  20  ft.  rise  is  fixed  at  the 
springings.     The  uniformly  distributed  load  upon  one-half  of  the  arch 
is  loo  tons,  and  upon  the  other  200  tons.     Find  the  bending  moment 
and  shearing  force  at  25  ft.  from  each  end. 

10.  An  arched  rib  with  parabolic  axis,  of  100  ft.  span  and  12^  ft.  rise, 
is  loaded  with  I  ton  at  the  centre  and  i  ton  at  20  ft.  from  the  centre, 
measured  horizontally.     Determine  the  thrusts  and  shears  along  the  rib 
at  the  latter  point,  and  show  how  they  will  be  affected  by  a  change  of 
100°  F.  from  the  mean  ;  the  coefficient  of  linear  expansion  being  .00125 
for  1 80°  F. 

u.  A  parabolic  arched  rib  hinged  at  the  ends,  of  64  ft.  span  and  16 
ft.  rise,  is  loaded  with  i  ton  at  each  of  the  points  of  division  of  eight 
equal  horizontal  divisions.  Find  the  horizontal  thrust  on  the  rib,  allowing 
for  a  change  of  60°  F.  from  the  mean  temperature.  Also  find  the  maxi- 
mum flange  stresses,  the  rib  being  of  double-tee  section  and  12  in.  deep 
throughout.  (Coefficient  of  linear  expansion  per  i°  F.  —  / -f-  144000.) 

12.  The  axis  of  an  arched  rib  of  50  ft.  span,  10  ft.  rise,  and  ninged  at 
both  ends  is  a  parabola.     Draw  the   linear  arch  when  the  rib  is  loaded 
with  two  weights  each  equal  to  2  tons  concentrated  at  two  points  10  ft. 
from  the  centre  of  the  span.     If  the  rib  is  of  double-tee  section  and  24 
in.  deep,  find  the  maximum  flange  stresses. 

If  the  arch  is  loaded  so  as  to  produce  a  stress  of  10,000  Ibs.  per  square 
inch  in  the  metal,  show  that  the  rib  will  deflect  .029  ft.,  E  being  25,000,000 
Ibs. 

13.  A  steel  parabolic  arched  rib  of  50  ft.  span  and  10  ft.  rise  is  hinged 
at  both  ends  and  loaded  at  the  centre  with  a  weight  of  12  tons.     Find 
the  horizontal  thrust  on  the  rib  when  the  temperature  varies  60°  F.  from 
the  mean,  and  also  find  the  maximum  flange  stresses,  the  rib  being  of 
double-tee  section  and  12  in.  deep. 

14.  A  semicircular  rib,  pivoted  at  the  crown  and  springings,  is  loaded 
uniformly  per  horizontal  unit  of  length.     Determine  the  position  and 
magnitude  of  the  maximum  bending  moments,  and  show  that  the  hori- 
zontal thrust  on  the  rib  is  one-fourth  of  the  total  load. 

15.  Draw  the  linear  arch  for  a  semicircular  rib  of  uniform  section 


EXAMPLES.  8ll 

under  a  load  uniformly  distributed  per  horizontal  unit  of  length  (a)  when 
hinged  at  both  ends  ;  (b)  when  hinged  at  both  ends  and  at  the  centre;  (c) 
when  fixed  at  both  ends. 

16.  A  semi-elliptic  rib  (axes  ia  and  2b)  is  pivoted  at  the  springing. 
Find  the  position  and  magnitude  of  the  maximum  bending  moment,  the 
load  being  uniformly  distributed  per  horizontal  unit  of  length. 

How  will  the  result  be  affected  if  the  rib  is  also  pivoted  at  the  crown  ? 

17.  Draw   the   equilibrium  polygon  for  a  parabolic    arch  of  100  ft. 
span  and  20  ft.  rise  when  loaded  with  weights  of  3,  2,  4,  and  2  tons,  re- 
spectively, at  the  end  of  the  third,  sixth,  eighth,  and  ninth  division  from 
the  left  support,  of  ten  equal  horizontal  divisions.     (Neglect  the  weight 
of  the  rib.) 

If  the  rib  consist  of  a  web  and  of  two  flanges  2^  ft.  from  centre  to 
centre,  determine  the  maximum  flange  stress. 

18.  Find  the  flange  stresses  at  the  ends  of  the  rib,  in  the  preceding 
question,  and  also  at  the  points  at  which  the  weights  are  concentrated, 
when  both  ends  are  absolutely  fixed. 

19.  A  semicircular  rib  of   28  ft.  span  carries  a  weight  of  £  ton  at  4  ft. 
(measured  horizontally)  from  the  centre.     Find  the  thrust  and  shear  at 
the  centre  of  the  rib  and  at  the   point  at  which  the  weight  is  concen- 
trated. 

20.  The  axis  of  an  arched  rib  hinged  at  both  ends,  for  a  span  of  50  ft. 
and  a  rise  of  10  ft.,  is  a  parabola.     Draw  the  equilibrium  polygon  when 
the  arch  is  loaded  with  two  equal  weights  of  2  tons  concentrated  at  two 
points  10  ft.  from  the  centre  of  the  span.     Also  determine  the  maximum 
flange  stress  in  the  rib,  which  is  a  double-tee  section  2  ft.  deep. 

21.  The  load  upon  a  parabolic  rib  of  50  ft.  span  and  15  ft.  rise,  hinged 
at  both  ends,  consists  of  weights  of  I,  2,  and  3  tons  at  points  15,  25,  and 
40  ft.,  respectively,  from  one  end.     Find  the  axial  thrusts  and  the  shears 
at  these  points. 

Ans.   Horizontal  thrust  =  9.6  tons. 

Axial  thrusts  :  above  i  ton  =  9.3  tons  ; 
below  i    "     =97     " 
above  3  tons  =  8.3     " 
below  3    "     =  io.r  " 

Shears  :  above  i  ton  =  3.  i  tons  ; 

below  i  "  =  2.2  " 
above  3  tons  =  5  " 
below  3  "  =  2.6  " 

22.  Draw  the  linear  arch  and  determine  the  maximum  flange  stresses 
for  an  arched  rib  of  80  ft.  span,  16  ft.  rise,  and  loaded  with  five  weights 
each  of  2  tons  at  the  end  of  the  first,  second,  third,  fourth,  and  fifth 
division,  of  eight  equal  horizontal  divisions.     The  rib  is  of  double-tee 


8l2  THEORY   OF  STRUCTURES. 

section  and  30  in.  deep.  Also  find  the  shears  and  the  axial  thrusts  at 
the  fifth  point  of  division. 

23.  A  wrought-iron    parabolic    lib  of   96  ft.  span   and   16  ft.  rise  is 
hinged  at   the   two   abutments;    it  is  of  a  double-tee  section  uniform 
throughout,  and  24  in.  deep  from  centre  to  centre  of  the  flanges.     Deter- 
mine the  compression  at  the  centre,  and  also  the  position  and  amount 
of  the  maximum  bending  moment  (a)  when  a  load  of  48  tons  is  concen- 
trated at  the  centre  ;  (b)  when  a  load  of  96  tons  is  uniformly  distributed 
per  horizontal  unit  of  length. 

Determine  the  deflection  of  the  rib  in  each  case. 

24.  Design  a  parabolic  arched  rib  of  100  ft.  span  and  20  It.  rise,  hinged 
at  both  ends  and  at  the  middle  joint  ;  dead  load  =40   tons  uniformly 
distributed  per  horizontal  unit  of  length,  and  live  load  =  i  ton  per  hori- 
zontal foot. 

25.  Show  how  the  calculations  in  the  preceding  question  are  affected 
when  both  ends  are  absolutely  fixed. 

26.  In  the  framed  arch  represented  by  the  figure,  the  span  is  120  ft., 

t*ie  r'Se  I2  *L>  l^e  c^ePt^1  °^ tne  truss  at  the 
crown  5  ft.,  the  fixed  load  at  each  top  joint 
FIG.  5o6.  10  tons,  and  the  moving  load  10  tons.  De- 

termine the  maximum  stress  in  each  member  with  any  distribution  of 
load.  Show  that,  approximately,  the  amount  of  metal  required  for  the 
arch  :  the  amount  required  for  a  bowstring  lattice-girder  of  the  same 
span  and  17  ft.  deep  at  the  centre  :  the  amount  required  for  a  girder  of 
the  same  span  and  12  ft.  deep  ::  100  :  155  :  175. 

27.  The  steel  parabolic  ribs  for  one  of  the  Harlem  River  bridges  has  a 
clear  opening  of  510  ft.,  a  rise  of  90  ft.,  a  depth  of  13  ft.,  and  are  spaced 
14  ft.  centre  to  centre.     The  dead  weight  per  lineal  foot  is  estimated  at 
33,000  Ibs.  and  the  live  load  at  8000  Ibs.  ;  a  variation  in  temperature  of 
75°  F.  from  the  mean  is  also  to  be  allowed  for.     Determine  the  maxi- 
mum bending  moment  (assuming  /constant),  and  the  maximum  deflec- 
tion.    R  =  26,000,000  Ibs.     Show  how  to  deduce  the  play  at  the  hinges. 

28.  A  cast-iron  arch  (see  figure)  whose  cross-sections  are  rectangular 
O'g'o'    »0 and  uniformly  3  in.  wide,  has  a  straight  horizon- 
tal extrados,  and  is  hinged  at  the  centre  and  at 
the  abutments.     Calculate  the  normal  intensity  of 
stress  at  the  top 'and  bottom  edges  D,  £  of  the 

FIG.  507.  vertical  section,  distant  5  ft.  from  the   centre  of 

the  span,  due  to  a  vertical  load  of  20  tons  concentrated  at  a  point  dis- 
tant 5  ft.  4  in.  horizontally  from  B.  Also  find  the  maximum  intensity 
of  the  shearing  stress  on  the  same  section,  and  state  the  point  at  which 
it  occurs.  (AB  —  21  ft.  4  in.). 


INDEX. 


Allowance  for  the  weight  of  a  beam,  403. 

Alternating  stresses,  152. 

American  iron  columns,  532. 

Anchorage,  704. 

Angle  of  repose,  237. 
"  torsion,  568. 

Angular  momentum,  177. 

Anti-friction  curve,  320. 
"  pivots,  320. 

Arch,  470. 

Arch  abutment,  maximum  thickness  of, 
649. 

Arch,  conditions  of  equilibrium  of,  745. 
"      formulae  for  thickness  of,  750. 
11      linear,  743,  750,  760. 

Arched  ribs,  740,  762. 

"         "      deflection  of,  780,  802. 

Arched  ribs,  effect  of   change    of    tem- 
perature on,  770,  786. 

Arched  ribs,  elementary  deformation  of, 
781. 

Arched  ribs,  general  equations  of  equi- 
librium of,  784. 

Arched  ribs,  graphical  determination  of 
stresses  in,  677. 

Arched    rib    of   uniform   stiffness,    788, 
789,  795,  800,  804. 

Arched  ribs  with  fixed  ends,  771. 
'*         "      with  hinged  ends,  764. 

Arched  ribs  with  axis  in  form  of  circu- 
lar arc,  769,  773. 

Arched    ribs   with    parabolic    axis,  760, 

775- 
Arched  ribs  with  semicircular  axis,  765, 

775- 

Arches,  middle-third  theory  of,  746. 
Auxiliary  truss,  719. 

Back-stays,  16,  704. 

Baker's  formulae  for  strength  of  pillars, 

549- 
Balancing,  198. 


Beam  acted  upon  by  oblique  forces,  396. 
Beam,  transverse  strength  of,  340,  429. 
Beam,  transverse  vibration  of  loaded, 

461. 
Beams,  equilibrium  of,  93. 

"        of  uniform  strength,  358-365. 
Bearing  surface,  314,  315. 
Belts,  321. 

"      effect  of  high  speed  in,  325. 

"      effective  tension  of,  324. 

"      slip  of,  326. 

"      stiffness  of,  327. 
Bending  moment,  96,  118,  434. 
Bending  moment  in  plane  which  is  not 

a  principal  plane,  354. 
Bending  moment,  relation  between,  and 

shearing  stress,  108. 
Bevel-wheels,  335. 
Boilers,  586. 
Bollman  truss,  56,  618. 
Bowstring  truss,  61,  618. 
Brace,  i,  25. 
Brakes,  323. 

Breaking-down  point,  149. 
Breaking  stress,  147. 

weights,  348,  399. 
Breaking  weights  of  iron  girders,  369, 

370.. 

Breaking  weights,  tables  of,  212. 
Brickwork,  1.49. 

Bridge,  bowsinng  suspension,  626. 
"        loads,  600. 
"        trusses,  17,  52. 

chords  of,  625. 
"         depth  of,  597. 
Bridge    trusses,     maximum     allowable 

stress  in,  657. 

Bridge  trusses,  stiffness  of,  598. 
"         stringers  of,  656. 
Bridges,  597. 

position  of  platform  of,  598. 
Buckling  of  pillars,  513,  515. 

813 


8i4 


INDEX. 


Cable  with  sloping  suspenders,  717. 
Cables,  703. 

'        curves  of,  706. 
'        deflection  of,  714. 
'        length  of  arc  of,  712. 
'        parameter  of,  711. 

weight  of,  713. 
Camber,  388,  659. 
Cantilever,  365. 

"  curve  of  boom  of,  634. 

deflection  of,  638. 
"  depth  of,  637. 

"  weight  of,  632. 

Cast-iron,  147. 
Catenary,  34,  706,  750. 
Cement,  150. 
Centres  of  gravity,  n. 
Centre  of  resistance,  I,  743. 
Centrifugal  force,  181. 
Centripetal  force,  182. 
Clapeyron's  theorem,  292. 
Coefficient  of  cubic  elasticity,  255. 
elasticity,   141,  143. 
fluidity,  162. 
hardness,  164. 
lateral  elasticity,  144. 
rigidity,  254,  285. 
rupture,  248. 
torsional  rupture,  574. 
transverse  elasticity,  285. 
Collar-beams,  25. 
Columns,  see  Pillars,  513,  538. 
flexure  of,  554,  5*57. 
Compound  strain,  236. 
Compression,  141. 
Conjugate  stresses,  247. 
Continuous  girders,  463. 
Continuous  girders,  advantages  and  dis- 
advantages of,  486. 
Continuous  girders,  maximum  bending 

moment  in,  465. 
Coulomb's  laws,  568. 
Counterbrace,  60. 
Counter-efficiency,  328. 
Counterforts,  270. 
Covers  of  riveted  joints,  665. 
Cranes,  13. 

bent,  31. 
"        derrick,  16. 
jib,  13- 
pit,  IS- 

Crank  effort,  207. 
Cubic  elasticity,  255. 
"      strain,   283. 

D<.ad  load,  143,  600. 
Deflection,  curve  of,  434. 

of  girders,  384-386,  638. 
Deformation,  140,  251,  254. 


Dock  walls,  270. 
Dynamometer,  Prony's,  327. 

Earth  foundations,  258. 
Earthwork,  255. 

pressure  of,  257. 
Earthwork,  Rankine's  theory  applied  to 

retaining  walls,  264. 
Efficiency  of  mechanisms,  335. 

"          of  riveted  joints,  666. 
Elastic  curve,  355. 

"       moment,  96,  340. 
Elasticity,  140. 

coefficient  of,  141,  143. 
"  cubic,  255. 
"  lateral,  144. 
"  transverse,  285, 
limit  of,  145. 
Ellipse  of  stress,  241. 
Ellipsoid  of  stress,  281. 
Empirical  rules  for  wind-pressure,  663. 
Encastr6  girders,  458. 
Energy,  207. 

"        curves  of,  207. 
"        fluctuation  of,  207. 
"        kinetic,  167,  169,  170. 
"         potential,  167. 
Envelope  of  moments,  121. 
Equalization  of  stress,  349. 
Equalizer,  629. 
Equilibrated  polygon,  740. 
Equilibrium  of  beams,  428. 
Equilibrium  of  beams,  general  equations 

of,  428. 

Equilibrium  of  flanged  girders,  366. 
Euler's  theory  of  the  strength  of  pillars, 

537- 
Examples,     69-92,     132-139,     216-234, 

294-298,    337-339*    407-427,  490-512, 

563-567,    580-585,   594~596>   689-702, 

734-739,  809-812. 
Expansion  of  solids,  215. 
Extension  of  prismatic  bar,  289. 
Extrados,  740. 
Eyebars,  661. 

steel,  665. 

Factor  of  safety,  150. 
Fatigue,  152. 
Fink  truss,  54. 
Flanged  girders,  365. 

"  "        equilibrium  of,  366. 

"        stiffness  of,  384. 
Flanges,  365,  597. 

"         curved,  366. 

"         horizontal.  366. 
Flexure    of    columns,   see    Pillars,   554, 

557- 
Flow  of  solids,  162. 


INDEX. 


815 


Fluctuation  of  stress,  151. 
Fluid  pressure,  162. 
Force  polygon,  3,  7,  119. 
Foundations,  earth,  258. 

limiting  depth  of,  258. 
of  walls,  270. 
Fracture,  141. 

Framed  arch,  stresses  in,  804. 
Framed  arch,  Clerk  Maxwell's  method 

of  determining  stresses  in  a,  806. 
Frames,  i,   2. 

incomplete,  27,  61. 
Friction,  300. 

angle  of,  237. 

coefficient  of,  300,  313. 

journal,  310. 

rolling,  310. 
Funicular  curve,  10. 

polygon,  3,  7,  117,  II9. 

Gins,  17. 

Girder  of  uniform  strength,  381. 

Gordon's  formulae  for  pillars,  522. 


Hinged  girders,  127-131. 
Hodgkinson's  formulae  for  the  strength 

of  pillars,  513.  517-521. 
Hooke's  law,  142. 
Howe  truss,  58,  611. 

Impact,  184. 
Impulse,  176. 
Incomplete  frames,  27. 
Inertia,  198. 

moment  of,  12,  342. 

pressure  due  to,  200. 
Inflection,  point  of,  453-463. 
Internal  stress,  235. 
Isotropic  bodies,  283. 

Joint  of  rupture,  747. 
Keystone,  741. 

Lateral  bracing,  654. 
Lattice  girder,  600. 
Launhardt's  formula,  153. 
Lenticular  truss,  626. 
Limit  of  elasticity,  145. 
Line  of  loads.  5. 

resistance,  273-276,  741,  750. 
"   rupture,  265. 
Linear  arch,  743,  753-760. 
Loads,  live,  in,  115,  IIQ>  600,639,641, 

Loads,  stationary  (dead),  118,  600. 
Long  pillars,  535. 

Mansard  roof,  6. 

Masonry,  149. 

Mechanical  advantage,  294. 


Middle  third  theory  of  arches,  746, 
Modulus  of  elasticity,  141. 

"  rupture,  348. 

"  transverse  elasticity,  254 
Moment  of  forces,  116. 

"  inertia,  12,  342. 

"        examples  of,  371-81. 
Moment  of  inertia,  variable  section   of 

•455- 
Moment  of  inflexibility,  96. 

"  resistance,  96. 
Momentum,  176. 
Mortar,  150. 

Neutral  axis,  340. 

of  a  loaded  beam,  435-454. 
surface,  340. 

Oblique  resistance,  169. 
OsciLatory  motion,   190,  195. 


Panel  points,  52. 
r^anels,  54. 
Diers,  65. 

513. 

Euler's  formulas  for,  527. 
failure  of,  515. 
flexure  of,  515. 

formulae  for  American,  527-532. 
Gordon's  formulae  for,  522 
Pillars,  Hodgkinson's  formulae  for  '517- 

521. 

Pillars,  Rankine's  formula  for,  526. 
Pillars  with  stress  uniformly  distributed 

516. 
Pillars   with   uniformly   varying   stress, 

Pins,  661. 

Piston  velocity,  curves  of,  205 

Pivots,  316. 

conical,  319 

cylindrica.,  316. 

Schiele's  (anti-friction),  320 
Plasticity,  141. 
Poisson's  ratio,  142 
Pole,  7. 
Polygon  of  forces,  3,  7. 

"  pressure,  743. 
Pratt  truss,  60. 
Primitive  strength,  153. 
Principals,  33,  34. 
Prony's  dynamometer,  327. 
Proof  strain,  171. 
stress.  171. 
Purchase    304. 
Purlins,  33,  34. 

Radius  of  gyration,  174,  528-531 
p"'°  of  twist,  289. 


8i6 


INDEX. 


Redundant  bars,  48. 

Reservoir  walls,  271. 

Resilience,  171. 

Retaining  walls,  260. 

Retaining  walls,   conditions   of   equiib- 

rium  of.  260. 

Retaining  walls,    Rankine's   theory   ap- 
plied to,  264. 
Rivet   connection    between   flange    and 

web,  660. 
Riveted  joints,  668. 

covers  of,  675. 
design  of,  678. 
efficiency  of,  679. 
failure  of,  670. 
stresses  in,  670. 
theory  of,  671-675. 
Riveting,  666. 
Rivets,  666. 

"        dimensions  of,  667. 
Rocker-link,  629. 
Rollers,  35,  639. 
Roof  trusses,  17. 

"         distribution  of  loads  on,  39. 
types  of,  33. 
weights  of,  37. 
Ropes,  321. 


Saddles,  704. 
Schiele's  pivots,  320. 
Screws,  306. 

"        endless,  309. 
Sections,  method  of,  62 
Set,  145.       -,.  '  *~~**v 
Shafting,  distance.  EetvySfen  bearings  of, 

575- '  •  •/'* *   1 

Shafting,  efficiency  of,  577. 

intprrfal-streiB&iii,  237. 
stiffness  of,  573. 

"    '     torsional  strength  of,  288-291. 
Shear,  141. 

"      maximum,  121,  237. 
Shearing  force,  95. 

"      '.  ^   et      examples  of,  97,  108. 
Shearing  force   and  bending    moment, 

relation  between,  108. 
Shearing  stress,  198. 

.  "        distribution  of,  391. 
Shear-legs,  17. 
Similar  girders,  401-404. 
Skew-backs,  34,  74°- 
Soffit,  740. 
Spandril,  740. 
Specific  weight,  143. 
Spherical  shells,  591. 
Spritigings,  740. 
Springs,  355,  456-458. 

"         simple  rectangular,  456. 


Springs,  spiral,  477. 

Springs  of  constant  depth,  but  triangu- 
lar in  plan,  4.57. 

Springs  of  constant  width,  but  parabolic 
in  elevation,  457. 

Statical  breaking  strength,  153. 

Steel,  148. 

Stiffening  truss,  719. 

"      hinged  at  centre,  725. 

Stiffness,  190,  387,  389. 

Strain,  140,  251. 

Straining  cill,  18. 

Stress,  141,  251 

"      and  strain,  relation  between,  281. 
"       general  equations  of,  277. 
principal,   241. 

planes  of,  237. 

Stresses,  conjugate,  248,  250. 

Stress-strain  curves,  147-149. 
line,  144. 

Strut,  i. 

St.  Venant's  torsion  results,  572. 

Surface  loading,  350. 

Suspenders,  706. 

Suspension-bridges,  703. 

"  loads  on,  730. 

Suspension-bridges,  modifications  of, 
73i. 

Suspension-bridges,  pressure  upon  piers 
of,  718. 

Swing-bridges,  470-472. 

Tables  of  breaking  weights  and  coeffi- 
cients of  bending  strength  of  timber, 
212,  213. 

Table   of   coefficients   of   axle  friction, 

336. 

Table  of  coefficients  in  Gordon's  for- 
mula, 524. 

Table  of  coefficients  in  Rankine's  mod- 
ification of  Gordon's  formula,  526. 

Tables  of  diagonal  and  chord  stresses, 
644-650. 

Tables  of  efficiencies,  587. 

"       "  elliptic  integrals,  562. 
"       "  expansion  of  solids,  215. 
"       "  eyebar  dimensions,  665. 
"       "  factors  of  safety,  214. 

Tables  of  loads  for  highway  bridges, 
687. 

Tables  of  strengths,  elasticities,  and 
weights  of  iron  and  steel,  210. 

Table  of  strengths,  elasticities,  and 
weights  of  various  alloys,  211. 

Tables  of  weights  of  modern  bridges, 
682-687. 

Table  of  weights  and  crushing  strength 
of  rocks,  214 

Table  of  weights  of  roof  coverings,  67. 


INDEX. 


8I7 


Table  of  weights  of  roof  frames,  67. 

Tension,  141. 

Theorem  of  three  moments,  463-470. 

Thick  hollow  cylinder,  588. 

Timber,  149. 

Torsion,  141,  568. 

"         St.  Venant's  results,  572. 
Torsional  coefficient  of  elasticity,  145. 

resilience,  574. 
Torsional  strength  of  shafts,  288,  289, 

571,  572. 

Transverse  strength,  141. 
Transverse  vibration  of  a  loaded  beam, 

461. 

Trellis  girder,  600. 

Tresca's  theory  of  flow  of  solids,  162. 
Tripods,  17. 
Truss,  2. 

'       composite,  31. 
king-post,  21. 

'       queen-post,  25. 

'       roof,  32. 

'        triangular,  19. 
Trussed  beams,  53. 
Twist,  141. 

Unwin's  formula,  159. 

Values  of  k*,  174,  528-531. 
Vibration  strength,  153. 
Voussoir,  741. 

Warren  truss,  57. 


Web  thickness,  382. 

Wedge,  303 

Weights  of  roof  coverings,  67. 

"    frames,  67. 
Weyrauch's  formula,  153. 
Weyrauch's  theory  of  buckling  of  pillars, 

550. 

Wheel  and  axle,  329. 
Whipple  truss,  618. 
Wind  pressure,  38,  67,  629,  651,  653. 
Wind-pressure,  American  specifications 

of,  652. 

Wind-Pressure  Commission  rules,  653. 
Wind-pressure,    empirical    regulations, 

653. 

Wohler's  law,  150. 
Work,   167. 

effective,  178. 
external,  168. 
internal,  168. 
useful,  178. 
waste,  178. 

Work  done  in  bending  a  beam,  460. 
Work  done  in   small   deformation  of  a 

body,  292. 

Work  of  journal  friction,  314. 
Working  load,  150. 

"         strength,  150. 
"         stress,  150. 
Wrought-iron,  148. 

Yield-point,  149. 


ERRATA. 

Page    12,  1  5th  line  from  top,  for  A\  ,  A*,  As,  read  a\  ,  a*,  a*. 
23,  Fig.  34,  for  BO  read  AO,  and  for  AO  read  BO. 

16562  16562 

47,  2d  line,  for       ,     read  —  -=-. 

52,  3d  line  from  bottom,  for  W*  read  W*. 
55,  8th  line  from  top,  for  2  7i  read  2  7\. 

63,  7th  line  from  bottom,  for  "revolving"  read  "  resolving." 

3  V  13  3  V's 

64,  3d  line  from  top,  for  -  —  read 

2V  13 


127,  Fig.   181,  line  x*  should  lie  between  W*  and  B  instead  of 

between  E  and  W*. 

215,  i  st  line,  for  ACTORS  read  FACTORS. 
219,  4th  line  from  top,  for  17,828  Ibs.  read  17,328  Ibs. 
223,  6th  line  from  top,  for  8625  ft.-lbs.  read  15,750  ft.-lbs. 
239,  nth  line  from  top,  in  2d  equation,  for  =  I  read  =  —  i. 
274,  5th  line  from  top,  for  CD  read  QP. 
276,  ist  line  (numerator),  for  ABQP  read  OBQ. 

294,  2d  line  from  bottom,  for  "thrust"  read  "tension,"  and  for 

"tension"  read  "thrust." 

295,  I2th  line  from  bottom,  for  11°  38'  read  3°  29';   nth  line  from 

bottom,  for  3°  26'  read  63°  26'. 

297,  loth  line  from  top,  for  9.8  ft.  read  12.2  ft.;  I3th  line  from 

bottom,  for  13.19  ft.  read  y  -  ft. 

298,  7th  line  from  bottom,  for  20,720  Ibs.  read  i4,o49TJt¥  Ibs. 
326,  3d  line  from  top,  for  $T*(<?a  —  i)  read  \T^a  —  i)v. 
344,  5th  line  from  bottom,  for  1000  Ibs.  read  100  Ibs. 

363,  for  equation  at  bottom  of  page  substitute 


WJ  WJ 

364,  8th  Iin6  from  top,  for read  - 

(Over) 


Page    371,  Fig.  292*2,  7/t  and  h*  should  be  interchanged. 
413,  I4th  line  from  top,  for  230  ft.  read  129.6  ft. 
417,  8th  line  from  top,  for  42^  tons  read  42  tons. 
422,  Add   to   last   line:    The   tensile   and    compressive   working 

strengths   being    2000    and    5000   Ibs.    per   square   inch, 

respectively.     Assume  that  the  thickness  of  the  web  is  a 

fraction  of  its  depth. 
"       423,  ist  line,  for  26^  in.  read  2o^G  in.,  and  for  28.8  sq.  in.  read 

25^  sq.  in.  ;  2d  line,  for  5.76  sq.  in.  read  6|f  sq.  in. ;  and 

3d  line,  for  93  sq.  read  53  sq. 

"       455,  4th  line  from  top,  equation,  for  —  read  —7. 

474,  i  ith  line  from  top,  for  "  the  wheels  of  a  locomotive  passing  " 

read  "  five  wheels  passing." 
485,  1 3th  line  from  top,  for/2  read  y*. 
"       501,  3d  line  from  top,  for  ^i  ,  fa  ,  fa  ,  read  p\ ,  pz ,  p<>. 

530,  ist  line  under  (»-),  for  S  =  26/1  —  I?  read  S  =  2.bh  —  &* ;  and 

in  denominator  of  next  equation,  for  //2  read  ^. 
555,  7th  line  from  bottom,  equation,  for  dQ  read  d(f>. 
565,  3d  line  of  Example  27,  for  "maximum"  read  "minimum." 
571,  bottom  line,  for  i.57/)2 Tread  1.57 D*Tf. 
578,  ist  line  of  Example,  for  50  read  95. 
.    "        580,  Examples  7  and  8,  for  m  read  G. 

581,  Examples  10,  u,  and  15,  for  m  read  G. 

582,  Examples  18  and  19,  for  m  read  G. 

583,  Example  30,  for  m  read  G. 

647,  Table  of  Maximum  Stresses  in  the  Verticals,  change  3d  and 
4th  lines  to  read 

7/3  =  42625  +  16200  =  58,825  Ibs. 
7/4  =  28700  +     5400  =  34,100    " 

649,  9th  line  from  top,  for  6250  read  7600,  and  for  1525  read  2875. 

650,  ist  line  of  Compression  Chord  Table,  for  di  read  d\. 
707,  Equation  (2),  for  PS  read  ps. 

711,  6th  line  from  bottom,  after  Art.  4  add  "  Case  B." 

714,  9th  line  from  top,  for  H\  sec  6  read  H  sec  6. 

715,  9th  line  from  bottom,  equation,  for  — ^  read  — . 
724,  4th  line  from  top,  for  iv  read  x. 

758,  Equation  7,  for  2iuyp  sin  —  read  2wyp\s\t\  — )  . 

u  I*  I 

772,  Equation  (5),  for  —  read  juE;  Equations  (6)  and  (7),  for  —^ 

read  nEI,  // 


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